Miscellaneous Exercise Class 12 Chapter 11​

Let OA be the line joining the origin \( (0,0,0) \) and the point \( \mathrm{A}(2,1,1) \).
Let BC be the line joining the points \( \mathrm{B}(3,5,-1) \) and \( \mathrm{C}(4,3,-1) \)
\( \text{Direction ratios of} \ O A=\left(a_{1}, b_{1}, c_{1}\right) \equiv[(2-0),(1-0),(1-0)] \equiv(2,1,1) \)
\( \text{Direction ratios of} \ BC=\left(a_{2}, b_{2}, c_{2}\right) \equiv[(4-3),(3-5),(-1+1)] \) \( \equiv(1,-2,0) \)
Given-
OA is \( \perp \) to BC
\( \therefore \) we have to prove that -
\( a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \)
\( \text {L.H.S }=a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=2 \times 1+1 \times(-2)+1 \times 0=2-2=0 \)
R.H.S \( =0 \)
Thus, L.H.S. \( = \text{R.H.S.} \ldots \text{PROVED}\)
Hence OA is \( \perp \) to BC .

Let \( l, \mathrm{~m}, \mathrm{n} \) be the direction cosines of the line perpendicular to each of the given lines. Then,
\(ll_{1}+\mathrm{mm}_{1}+\mathrm{nn}_{1}=0\ldots (1)\)
and, \( ll_{2}+\mathrm{mm}_{2}+\mathrm{nn}_{2}=0 \ldots (2)\)
On solving (1) and (2) by cross - multiplication, we get -
\(\frac{l}{m_{1} n_{2}-m_{2} n_{1}}=\frac{m}{n_{1} l_{2}-n_{2} l_{1}}=\frac{n}{l_{1} m_{2}-l_{2} m_{1}}\)
Thus, the direction cosines of the given line are proportional to
\(\left(\mathrm{m}_{1} \mathrm{n}_{2}-\mathrm{m}_{2} \mathrm{n}_{1}\right),\left(\mathrm{n}_{1} l_{2}-\mathrm{n}_{2} l_{1}\right),\left(l_{1} \mathrm{~m}_{2}-l_{2} \mathrm{~m}_{1}\right)\)
So, its direction cosines are
\(\frac{m_{1} n_{2}-m_{2} n_{1}}{\lambda}, \frac{n_{1} l_{2}-n_{2} l_{1}}{\lambda}, \frac{l_{1} m_{2}-l_{2} m_{1}}{\lambda}\)
Where \( \lambda=\sqrt{\left(m_{1} n_{2}-m_{2} n_{1}\right)^{2}+\left(n_{1} l_{2}-n_{2} l_{1}\right)^{2}+\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}} \)
we know that -
\(\left(l_{1}^{2}+m_{1}^{2}+n_{1}^{2}\right)\left(l_{2}^{2}+m_{2}^{2}+n_{2}^{2}\right)-\left(l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right)^{2}\)
\(=\left(m_{1} n_{2}-m_{2} n_{1}\right)^{2}+\left(n_{1} l_{2}-n_{2} l_{1}\right)^{2}+\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}\ldots (3)\)
It is given that the given lines are perpendicular to each other. Therefore, \( l_{1} l_{2}+\mathrm{m}_{1} \mathrm{~m}_{2}+\mathrm{n}_{1} \mathrm{n}_{2}=0 \)
Also, we have
\(l_{1}^{2}+\mathrm{m}_{1}^{2}+\mathrm{n}_{1}^{2}=1\)
and, \( l_{2}^{2}+\mathrm{m}_{2}^{2}+\mathrm{n}_{2}^{2}=1 \)
Putting these values in (3), we get -
\(\left(\mathrm{m}_{1} \mathrm{n}_{2}-\mathrm{m}_{2} \mathrm{n}_{1}\right)^{2}+\left(\mathrm{n}_{1} l_{2}-\mathrm{n}_{2} l_{1}\right)^{2}+\left(l_{1} \mathrm{~m}_{2}-l_{2} \mathrm{~m}_{1}\right)^{2}=1\)
\(\Rightarrow \lambda=1\)
Hence, the direction cosines of the given line are \( \left(m_{1} n_{2}-m_{2} n_{1}\right),\left(n_{1} l_{2}-\right. \) \( \left.\mathrm{n}_{2} l_{1}\right),\left(l_{1} \mathrm{~m}_{2}-l_{2} \mathrm{~m}_{1}\right) \)

Angle between the lines with direction ratios \( a_{1}, b_{1}, c_{1} \) and \( a_{2}, b_{2}, c_{2} \) is given by
\(\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right|\)
Given -
\(\mathrm{a}_{1}=\mathrm{a}, \mathrm{b}_{1}=\mathrm{b}, \mathrm{c}_{1}=\mathrm{c}\)
\(\mathrm{a}_{2}=\mathrm{b}-\mathrm{c}, \mathrm{b}_{2}=\mathrm{c}-\mathrm{a}, \mathrm{c}_{2}=\mathrm{a}-\mathrm{b}\)
So,
\(\cos \theta=\left|\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}}\right|\)
\(=0\)
\(\therefore \cos \theta=0\)
So, \( \theta=90^{\circ} \)
Hence, Angle between the given pair of Lines is \( 90^{\circ} \).

Equation of a line passing through \( \left(x_{1}, y_{1}, z_{1}\right) \) and parallel to a line with direction ratios \( \mathrm{a}, \mathrm{b}, \mathrm{c} \) is
\(\frac{x-x_{1}}{\mathrm{a}}=\frac{y-y_{1}}{\mathrm{~b}}=\frac{z-z_{1}}{\mathrm{c}}\)
Since the line passes through origin i.e. \( (0,0,0) \)
\(x_{1}=0, y_{1}=0, z_{1}=0\)
Since line is parallel to x - axis,
\(a=1, b=0, c=0\)
Equation of Line is given by -
\(\Rightarrow \frac{x-0}{0}=\frac{y-0}{0}=\frac{z-0}{0}\)
\(\Rightarrow \frac{x}{1}=\frac{y}{0}=0\)

Angle between the lines with direction ratios \( a_{1}, b_{1}, c_{1} \) and \( a_{2}, b_{2}, c_{2} \) is given by
\(\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right|\)
A line passing through \( \mathrm{A}\left(x_{1}, y_{1}, z_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}, z_{2}\right) \) has direction ratios \( \left(x_{1}-x_{2}\right),\left(y_{1}-y_{2}\right),\left(z_{1}-z_{2}\right) \)
Direction ratios of line joining the points \( \mathrm{A}(1,2,3) \) and \( \mathrm{B}(4,5,7) \)
\(=(4-1),(5-2),(7-3)\)
\(=(3,3,4)\)
\(\therefore \mathrm{a}_{1}=3, \mathrm{~b}_{1}=3, \mathrm{c}_{1}=4\)
Direction ratios of line joining the points \( \mathrm{C}(-4,3,-6) \) and \( \mathrm{B}(2,9,2) \)
\(=(2-(-4)),(9-3),(2-(-6))\)
\(=(6,6,8)\)
\(\therefore \mathrm{a}_{2}=6, \mathrm{~b}_{2}=6, \mathrm{c}_{2}=8\)
Now,
\(\Rightarrow \cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right|\)
\(\Rightarrow \cos \theta=\left|\frac{3 \times 6+3 \times 6+4 \times 8}{\sqrt{3^{2}+3^{2}+4^{2}} \sqrt{6^{2}+6^{2}+8^{2}}}\right|\)
\(=\left|\frac{18+18+32}{\sqrt{9+9+16} \sqrt{36+36+64}}\right|\)
\(=\left|\frac{68}{\sqrt{34} \sqrt{4 \times 34}}\right|\)
\(=\left|\frac{68}{34 \times 2}\right| \therefore \cos \theta=1\)
So, \( \theta=0^{\circ} \)
Hence, Angle between the lines AB and CD is \( 0^{\circ} \).

Two lines \( \frac{x-x_{1}}{\mathrm{a}_{1}}=\frac{y-y_{1}}{\mathrm{~b}_{1}}=\frac{z-z_{1}}{\mathrm{c}_{1}} \) and \( \frac{x-x_{2}}{\mathrm{a}_{2}}=\frac{y-y_{2}}{\mathrm{~b}_{2}}=\frac{z-z_{2}}{\mathrm{c}_{2}} \) is are perpendicular to each other if
\(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)
Given -
\(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\)
comparing with
\(\Rightarrow \frac{x-x_{1}}{\mathrm{a}_{1}}=\frac{y-y_{1}}{\mathrm{~b}_{1}}=\frac{z-z_{1}}{\mathrm{c}_{1}}\)
We get -
\(\begin{array}{l}
x_{1}=1, y_{1}=2, z_{1}=3 \\
\& \ \mathrm{a}_{1}=-3, \mathrm{~b}_{1}=2 \mathrm{k}, \mathrm{c}_{1}=2
\end{array}\)
Similarly,
\(\Rightarrow \frac{x-1}{3 k}=\frac{y-2}{1}=\frac{z-3}{-5}\)
comparing with
\(\Rightarrow \frac{x-x_{2}}{\mathrm{a}_{2}}=\frac{y-y_{2}}{\mathrm{~b}_{2}}=\frac{z-z_{2}}{\mathrm{c}_{2}}\)
we get -
\(x_{2}=1, y_{2}=2, z_{2}=3\)
\( \& \ \mathrm{a}_{2}=3 \mathrm{k}, \mathrm{b}_{2}=1, \mathrm{c}_{2}=-5 \)
Since the two lines are perpendicular,
\(\mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}=0\)
\(\Rightarrow(-3) \times 3 \mathrm{k}+2 \mathrm{k} \times 1+2 \times(-5)=0\)
\(\Rightarrow-9 \mathrm{k}+2 \mathrm{k}-10=0\)
\(\Rightarrow-7 \mathrm{k}=10\)
\(\therefore \mathrm{k}=-\frac{10}{7}\)
Hence, the value of k is \( -\frac{10}{7} \).

The vector equation of a line passing through a point with position vector \( \overrightarrow{a} \) and parallel to vector \( \overrightarrow{b} \) is
\(\Rightarrow \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\)
Given, the line passes through \( (1,2,3) \)
So, \( \overrightarrow{a}=1 \hat{i}+2 \hat{j}+3 \hat{k} \)
Finding normal of plane
\(\overrightarrow{\mathrm{r}}(\hat{i}+2 \hat{j}-5 \hat{\mathrm{k}})+9=0\)
\(\Rightarrow \overrightarrow{\mathrm{r}} \cdot(\hat{i}+2 \hat{j}-5 \hat{\mathrm{k}})=-9\)
\(\Rightarrow-\overrightarrow{\mathrm{r}}(\hat{i}+2 \hat{j}-5 \hat{\mathrm{k}})=9\)
\(=\overrightarrow{\mathrm{r}}(-1 \hat{i}+2 \hat{j}+3 \hat{k})+9=0\)
Comparing with \( \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}}=\mathrm{d} \)
\(\overrightarrow{n}=-\hat{i}-2 \hat{j}+5 \hat{k}\)
Since line is perpendicular to plane, the line will be parallel of the plane
\(\therefore \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{n}}=-\hat{i}-2 \hat{j}+5 \hat{\mathrm{k}}\)
Hence,
\(\overrightarrow{r} =(1 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-\hat{i}-2 \hat{j}+5 \hat{k})\)
\(\overrightarrow{r} =(1 \hat{i}+2 \hat{j}+3 \hat{k})-\lambda(\hat{i}+2 \hat{j}-5 \hat{k})\)
This is the required vector equation of line.

The equation of a plane passing through \( \left(x_{1}, y_{1}, z_{1}\right) \) and perpendicular to a line with direction ratios \( \mathrm{A}, \mathrm{B}, \mathrm{C} \) is
\( \mathrm{A}\left(x-x_{1}\right)+\mathrm{B}\left(y-y_{1}\right)+\mathrm{C}\left(z-z_{1}\right)=0 \)
The plane passes through \( (a, b, c) \)
So, \( x_{1}=\mathrm{a}, y_{1}=\mathrm{b}, z_{1}=\mathrm{c} \)
Since both planes are parallel to each other, their normal will be parallel
\( \therefore \) Direction ratios of normal
\( = \) Direction ratios of normal of \( \overrightarrow{r} .(\hat{i}+\hat{j}+\hat{k})=2 \)
Direction ratios of normal \( =(1,1,1) \)
\(\therefore \mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=1\)
Thus,
Equation of plane in cartesian form is
\(\mathrm{A}\left(x-x_{1}\right)+\mathrm{B}\left(y-y_{1}\right)+\mathrm{C}\left(z-z_{1}\right)=0\)
\(\Rightarrow 1(x-\mathrm{a})+1(y-\mathrm{b})+1(z-\mathrm{c})=0\)
\(\Rightarrow x+y+z-(\mathrm{a}+\mathrm{b}+\mathrm{c})=0\)
Thus, \( x+y+z=\mathrm{a}+\mathrm{b}+\mathrm{c} \) is the required equation of plane.

Shortest distance between lines with vector equations \( \overrightarrow{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \) and \( \overrightarrow{r}=\overrightarrow{a_{2}}+\lambda \overrightarrow{b_{2}} \) is
\(\Rightarrow\left|\frac{\left(\overrightarrow{b_{1}} \times {\overrightarrow{b}_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|\)
Given -
\(\overrightarrow{r}=(6 \hat{i}+2 \hat{j}+2 \hat{k})+\lambda(1 \hat{i}-2 \hat{j}+\hat{2})\)
Comparing with \( \overrightarrow{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \), we get -
\(\overrightarrow{a_{1}}=(6 \hat{i}+2 \hat{j}+2 \hat{k}) \& \ \overrightarrow{b_{1}}=(1 \hat{i}-2 \hat{j}+\hat{2})\)
Similarly,
\(\overrightarrow{\mathrm{r}}=(-4 \hat{i}-\hat{\mathrm{k}})+\mu(3 \hat{i}-2 \hat{j}-2 \hat{\mathrm{k}})\)
Comparing with \( \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}_{2}}+\lambda \overrightarrow{\mathrm{b}_{2}} \), we get -
\(\overrightarrow{a_{1}}=(-4 \hat{i}-\hat{k}) \& \ \overrightarrow{b_{1}}=(3 \hat{i}-2 \hat{j}-2 \hat{k})\)
Now,
\(\Rightarrow\left(\overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}\right)=(-4 \hat{i}-\hat{\mathrm{k}})-(6 \hat{i}+2 \hat{j}+2 \hat{\mathrm{k}}) \\
=((-4-6) \hat{i}+(0-2) \hat{j}+(-1-2) \hat{\mathrm{k}}) \\
=(-10 \hat{i}-2 \hat{j}-3 \hat{\mathrm{k}}) \)
\(\begin{array}{l} \Rightarrow\left(\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right)=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 2 \\
3 & -2 & -2
\end{array}\right| \end{array}\)
\(\Rightarrow=\hat{i}[(-2 \times-2)-(-2 \times 2)]-\hat{j}[(1 \times-2)-(3 \times 2)] \hat{k}[(1 \times-2)-(3x \times 2)] \)
\(\Rightarrow=\hat{i}[4+4]-\hat{j}[-2-6]+\hat{\mathrm{k}}[-2+6] \)
\(\Rightarrow=8 \hat{i}+8 \hat{j}+4 \hat{\mathrm{k}}\)
\( \text{Magnitude of }\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}=\left|\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right|=\sqrt{8^{2}+8^{2}+4^{2}}=\sqrt{64+64+16} \)
\(=\sqrt{144}\)
\(=12\)
Also,
\(\Rightarrow\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(8 \hat{i}+8 \hat{j}+4 \hat{k}) \cdot(-10 \hat{i}-2 \hat{j}-3 \hat{k})\)
\(=-80+(-16)+(-12)\)
\(=-108\)
Shortest Distance \( =\left|\frac{\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|=\left|\frac{-108}{12}\right|=|-9|=9 \)
Hence, the shortest distance between the given two lines is 9 .

The vector equation of a line passing through two points with position vectors \( \overrightarrow{a} \& \overrightarrow{b} \) is
\(\Rightarrow \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})\)
The position vector of point \( \mathrm{A}(5,1,6) \) is given as -
\(\Rightarrow \overrightarrow{a}=5 \hat{i}+\hat{j}+6 \hat{k}\)
The position vector of point B \( (3,4,1) \) is given as -
\(\Rightarrow \overrightarrow{\mathrm{b}}=3 \hat{i}+4 \hat{j}+\hat{k}\)
\(\Rightarrow(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})=(3 \hat{i}+4 \hat{j}+\hat{k})-(5 \hat{i}+\hat{j}+6 \hat{k})\)
\(=(3-5) \hat{i}+(4-1) \hat{j}+(1-6) \hat{k}\)
\(=(-2 \hat{i}+3 \hat{j}-5 \hat{k})\)
\(\therefore \overrightarrow{\mathrm{r}}=(5 \hat{i}+\hat{j}+6 \hat{k})+\lambda(-2 \hat{i}+3 \hat{j}-5 \hat{k})\ldots (1)\)
Let the coordinates of the point where the line crosses the \(YZ\) plane be \( (0 \), \( y, z \) )
So, \( \overrightarrow{\mathrm{r}}=(0 \hat{i}+y \hat{j}+z \hat{\mathrm{k}}) \ldots (2)\)
Since point lies in line, it will satisfy its equation,
Putting (2) in (1)
\((0 \hat{i}+y \hat{j}+z \hat{\mathrm{k}})=(5 \hat{i}+\hat{j}+6 \hat{\mathrm{k}})+\lambda(-2 \hat{i}+3 \hat{j}-5 \hat{\mathrm{k}})\)
\(\Rightarrow(0 \hat{i}+y \hat{j}+z \hat{\mathrm{k}})=(5-2 \lambda) \hat{i}+(1+3 \lambda) \hat{j}+(6-5 \lambda) \hat{\mathrm{k}}\)
Two vectors are equal if their corresponding components are equal
So,
\(0=5-2 \lambda\ldots (3)\)
\(y=1+3 \lambda \ldots (4)\)
and, \( z=6-5 \lambda \ldots (5)\)
From equation (3), we get -
\(\lambda=\frac{5}{2}\)
Substitute the value of \( \lambda \) in equation (4) and (5), we get -
\(y=1+3 \lambda=1+3 \times(\frac{5}{2})=1+(\frac{15}{2})=\frac{17}{2}\)
and
\(z=6-5 \lambda=6-5 \times(\frac{5}{2})=6-(\frac{25}{2})=-\frac{13}{2}\)
Therefore, the coordinates of the required point are \( (0,\frac{17}{2},-\frac{13}{2}) \).

The vector equation of a line passing through two points with position vectors \( \overrightarrow{a} \& \overrightarrow{b} \) is
\(\Rightarrow \overrightarrow{r}=\overrightarrow{a}+\lambda(\overrightarrow{b}-\overrightarrow{a})\)
The position vector of point \( \mathrm{A}(5,1,6) \) is given as -
\(\Rightarrow \overrightarrow{a}=5 \hat{i}+\hat{j}+6 \hat{k}\)
The position vector of point \( B(3,4,1) \) is given as -
\(\Rightarrow \overrightarrow{\mathrm{b}}=3 \hat{i}+4 \hat{j}+\hat{\mathrm{k}}\)
\(\Rightarrow(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})=(3 \hat{i}+4 \hat{j}+\hat{\mathrm{k}})-(5 \hat{i}+\hat{j}+6 \hat{\mathrm{k}})\)
\(=(3-5) \hat{i}+(4-1) \hat{j}+(1-6) \hat{k}\)
\(=(-2 \hat{i}+3 \hat{j}-5 \hat{k})\)
\(\therefore \overrightarrow{\mathrm{r}}=(5 \hat{i}+\hat{j}+6 \hat{k})+\lambda(-2 \hat{i}+3 \hat{j}-5 \hat{k})\ldots (1)\)
Let the coordinates of the point where the line crosses the YZ plane be \( (0 \), \( y, z) \)
So, \( \overrightarrow{\mathrm{r}}=(x \hat{i}+0 \hat{j}+z \hat{\mathbf{k}}) \ldots (2)\)
Since point lies in line, it will satisfy its equation,
Putting (2) in (1)
\((x \hat{i}+0 \hat{j}+z \hat{k})=(5 \hat{i}+\hat{j}+6 \hat{\mathrm{k}})+\lambda(-2 \hat{i}+3 \hat{j}-5 \hat{\mathrm{k}})\)
\(\Rightarrow(x \hat{i}+0 \hat{j}+z \hat{\mathrm{k}})=(5-2 \lambda) \hat{i}+(1+3 \lambda) \hat{j}+(6-5 \lambda) \hat{\mathrm{k}}\)
Two vectors are equal if their corresponding components are equal
So,
\( x=5-2 \lambda \ldots (3)\)
\( 0=1+3 \lambda \ldots (4)\)
and, \( z=6-5 \lambda \ldots (5)\)
From equation (3), we get -
\( \lambda=-\frac{1}{3}\)
Substitute the value of \( \lambda \) in equation (4) and (5), we get \( x=5-2 \lambda=5-2 \times(-\frac{1}{3})=5+(\frac{2}{3})=\frac{17}{3} \)
and
\(z=6-5 \lambda=6-5 \times(-\frac{1}{3})=6+(\frac{5}{3})=\frac{23}{3}\)
Therefore, the coordinates of the required point are \( (\frac{17}{3},0,\frac{23}{3}) \).

The equation of a line passing through two points \( \mathrm{A}\left(x_{1}, y_{1}, z_{1}\right) \) and \( \mathrm{B}\left(x_{2}\right. \), \( \left.y_{2}, z_{2}\right) \) is
\(\Rightarrow \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\)
Given the line passes through the points A \( (3,-4,-5) \) and
B \( (2,-3,1) \)
\(\therefore x_{1}=3, y_{1}=-4, z_{1}=-5\)
and, \( x_{2}=2, y_{2}=-3, z_{2}=1 \)
So, the equation of line is
\(\Rightarrow \frac{x-3}{2-3}=\frac{y-(-4)}{-3-(-4)}=\frac{z-(-5)}{1-(-5)}\)
\(\Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=k\)
So,
\(x=-\mathrm{k}+3|y=\mathrm{k}-4| z=6 \mathrm{k}-5\ldots (1)\)
Let \( (x, y, z) \) be the coordinates of the point where the line crosses the plane
\(2 x+y+z+7=0\)
Putting the value of \( x, y, z \) from (1) in the equation of plane,
\(2 x+y+z+7=0\)
\(\Rightarrow 2(-\mathrm{k}+3)+(\mathrm{k}-4)+(6 \mathrm{k}-5)=7\)
\(\Rightarrow 5 \mathrm{k}-3=7\)
\(\Rightarrow 5 \mathrm{k}=10\)
\(\therefore \mathrm{k}=2\)
Putting the value of k in \( x, y, z \)
\(x=-\mathrm{k}+3=-2+3=1\)
\(y=\mathrm{k}-4=2-4=-2\)
\(z=6 \mathrm{k}-5=12-5=7\)
Hence, the coordinates of the required point are \( (1,-2,7) \).

The equation of a plane passing through \( \left(x_{1}, y_{1}, z_{1}\right) \) is given by
\( \mathrm{A}\left(x-x_{1}\right)+\mathrm{B}\left(y-y_{1}\right)+\mathrm{C}\left(z-z_{1}\right)=0 \)
where, \( \mathrm{A}, \mathrm{B}, \mathrm{C} \) are the direction ratios of normal to the plane.
Now the plane passes through \( (-1,3,2) \)
So, equation of plane is
\(A(x+1)+B(y-3)+C(z-2)=0\ldots(1)\)
Since this plane is perpendicular to the given two planes.
So, their normal to the plane would be perpendicular to normal of both planes.
we know that -
\( \overrightarrow{a} \times \overrightarrow{b} \) is perpendicular to both \( \overrightarrow{a} \& \overrightarrow{b} \)
So, required normal is cross product of normal of planes
\(x+2 y+3 z=5 \text { and } 3 x+3 y+z=0 \)
\(\begin{array}{l} \text { required Normal }=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{j} & \hat{\mathrm{k}} \\
1 & 2 & 3 \\
3 & 3 & 1
\end{array}\right| \end{array}\)
\(=\hat{i}[2(1)-3(3)]-\hat{j}[1(1)-3(3)]+\hat{\mathrm{k}}[1(3)-3(2)] \)
\(=\hat{i}[2-9]-\hat{j}[1-9]+\hat{\mathrm{k}}[3-6]\)
\(=-7 \hat{i}+8 \hat{j}-3 \hat{\mathrm{k}}\)
Hence, direction ratios \( =-7,8,-3 \)
\(\therefore A=-7, B=8, C=-3\)
Putting above values in (1), we get -
\(\begin{array}{l}
\mathrm{A}(\mathrm{x}+1)+\mathrm{B}(\mathrm{y}-3)+\mathrm{C}(\mathrm{z}-2)=0 \\
\Rightarrow-7(x+1)+8(y-3)+(-3)(\mathrm{z}-2)=0 \\
\Rightarrow-7 x-7+8 y-24-3 z+6=0
\end{array}\)
\(\begin{array}{l}
\Rightarrow-7 x+8 y-3 z-25=0 \\
\therefore 7 x-8 y+3 z+25=0
\end{array}\)
Therefore, equation of the required plane is \( 7 x-8 y+3 z+25=0 \).

The distance of a point with position vector \( \overrightarrow{a} \) from the plane \( \overrightarrow{r} \cdot \overrightarrow{n}=d \) is \( \left|\frac{\overrightarrow{a} \cdot \overrightarrow{\mathrm{n}}-\mathrm{d}}{|\overrightarrow{\mathrm{n}}|}\right| \).
The position vector of point \( (1,1, p) \) is given as -
\(\Rightarrow \overrightarrow{a_{1}}=1 \hat{i}+1 \hat{j}-p \hat{k}\)
The position vector of point \( (-3,0,1) \) is given as -
\(\Rightarrow \overrightarrow{\mathrm{a}_{2}}=-3 \hat{i}+0 \hat{j}+1 \hat{\mathrm{k}}\)
It is given that the points \( (1,1, \mathrm{p}) \) and \( (-3,0,1) \) are equidistant from the plane
\(\overrightarrow{\mathrm{r}} \cdot(3 \hat{i}+4 \hat{j}-12 \hat{\mathrm{k}})+13=0\)
\(\therefore\left|\frac{(1 \hat{i}+1 \hat{j}-\mathrm{p} \hat{k}) \cdot(3 \hat{i}+4 \hat{j}-12 \mathrm{k})+13}{\sqrt{3^{2}+4^{2}+(-12)^{2}}}\right|=\left|\frac{(-3 \hat{i}+0 \hat{j}+1 \hat{k}) \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})+13}{\sqrt{3^{2}+4^{2}+(-12)^{2}}}\right|\)
\(\Rightarrow\left|\frac{3+4-12 \mathrm{p}+13}{\sqrt{9+16+44}}\right|=\left|\frac{-9+0-12+13}{\sqrt{9+16+44}}\right|\)
\(\Rightarrow\left|\frac{20-12 \mathrm{p}}{\sqrt{169}}\right|=\left|\frac{-8}{\sqrt{169}}\right|\)
\(\Rightarrow 20-12 \mathrm{p}= \pm 8\)
\(\Rightarrow 20-12 \mathrm{p}=8 \text { or, } 20-12 \mathrm{p}=-8\)
\(\Rightarrow 12 \mathrm{p}=12 \text { or, } 12 \mathrm{p}=28\)
\(\therefore \mathrm{p}=1 \text { or, } \mathrm{p}=\frac{ 7 }{ 3 }\)
us, the possible values of \( p \) are 1 and \( \frac{ 7 }{ 3 } \).

The equation of any plane through the line of intersection of the planes \( \overrightarrow{r} . \overrightarrow{n_{1}}=d_{1} \) and \( \overrightarrow{r} . \overrightarrow{n_{2}}=d_{2} \) is given by -
\(\overrightarrow{r} \cdot \overrightarrow{n_{1}}=d_{1}+\lambda\left(\overrightarrow{r} \cdot \overrightarrow{n_{2}}=d_{2}\right)=0\)
So, the equation of any plane through the line of intersection of the given planes is
\({[\overrightarrow{\mathrm{r}} \cdot(\hat{i}+\hat{j}+\hat{\mathrm{k}})-1]+\lambda[\overrightarrow{\mathrm{r}} \cdot(2 \hat{i}+3 \hat{j}-\hat{\mathrm{k}})-4]=0}\)
\(\overrightarrow{\mathrm{r}} \cdot((1-2 \lambda) \hat{i}+(1-3 \lambda) \hat{j}+(1+\lambda) \hat{\mathrm{k}})-1-4 \lambda=0\)
\(\therefore \overrightarrow{\mathrm{r}} \cdot((1-2 \lambda) \hat{i}+(1-3 \lambda) \hat{j}+(1+\lambda) \hat{\mathrm{k}})=1-4 \lambda \ldots(1)\)
Since this plane is parallel to \(x\) -axis.
So, the normal vector of the plane (1) will be perpendicular to x -axis.
\( \text{Direction ratios of Normal}\left(a_{1}, b_{1}, c_{1}\right) \equiv[(1-2 \lambda),(1-3 \lambda),(1+)] \)
Direction ratios of x -axis \( \left(\mathrm{a}_{2}, \mathrm{~b}_{2}, \mathrm{c}_{2}\right) \equiv(1,0,0) \)
Since the two lines are perpendicular,
\(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)
\((1-2 \lambda) \times 1+(1-3 \lambda) \times 0+(1+\lambda) \times 0=0\)
\(\Rightarrow(1-2 \lambda)=0\)
\(\therefore \lambda=\frac{ 1 }{ 2 }\)
Putting the value of \( \lambda \) in (1), we get -
\(\Rightarrow \overrightarrow{r} \cdot((1-2 \lambda) \hat{i}+(1-3 \lambda) \hat{j}+(1+\lambda) \hat{k})=1+4 \lambda\)
\(\overrightarrow{r} \cdot\left(\left(1-2\left(\frac{1}{2}\right)\right) \hat{i}+\left(1-3\left(\frac{1}{2}\right)\right) \hat{j}+\left(1+\left(\frac{1}{2}\right)\right) \hat{k}\right)=1-4\left(\frac{1}{2}\right)\)
\(\Rightarrow \overrightarrow{r} \cdot(0 \hat{i}-\hat{j}+3 \hat{k})=6\)
Hence, the equation of the required plane is
\(\overrightarrow{r} \cdot(0 \hat{i}-\hat{j}+3 \hat{k})=6\)

The equation of a plane passing through \( \left(x_{1}, y_{1}, z_{1}\right) \) and perpendicular to a line with direction ratios \( \mathrm{A}, \mathrm{B}, \mathrm{C} \) is
\(\mathrm{A}\left(x-x_{1}\right)+\mathrm{B}\left(y-y_{1}\right)+\mathrm{C}\left(z-z_{1}\right)=0\)
The plane passes through \( \mathrm{P}(1,2,3) \)
So, \( x_{1}=1, y_{1}=2, z_{1}=-3 \)
Normal vector to plane \( =\overrightarrow{\mathrm{OP}} \)
where \( \mathrm{O}(0,0,0), \mathrm{P}(1,2,-3) \)
Direction ratios of \( \overrightarrow{\mathrm{OP}}=(1-0),(2-0),(-3-0) \)
\(=(1,2,-3)\)
\(\therefore A=1, B=2, C=-3\)
Equation of plane in cartesian form is
\(1(\mathrm{x}-1)+2(\mathrm{y}-2)-3(\mathrm{z}-(-3))=0\)
\(\Rightarrow \mathrm{x}-1+2 \mathrm{y}-4-3 \mathrm{z}-9=0\)
\(\Rightarrow \mathrm{x}+2 \mathrm{y}-3 \mathrm{z}-14=0\)

The equation of any plane through the line of intersection of the planes \( \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}_{1}}=\mathrm{d}_{1} \) and \( \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}_{2}}=\mathrm{d}_{2} \) is given by -
\(\Rightarrow\left(\overrightarrow{r} \cdot \overrightarrow{n_{1}}=d_{1}\right)+\lambda\left(\overrightarrow{r} \cdot \overrightarrow{n_{2}}=d_{2}\right)=0\)
So, the equation of any plane through the line of intersection of the given planes is
\({[\overrightarrow{\mathrm{r}} \cdot(\hat{i}+2 \hat{j}+3 \hat{\mathrm{k}})-4]+\lambda[\overrightarrow{\mathrm{r}} \cdot(-2 \hat{i}-\hat{j}+\hat{\mathrm{k}})-5]=0}\)
\(\overrightarrow{\mathrm{r}} \cdot((1-2 \lambda) \hat{i}+(2-\lambda) \hat{j}+(3+\lambda) \hat{\mathrm{k}})-4-5 \lambda=0\)
\(\therefore \overrightarrow{\mathrm{r}} \cdot((1-2 \lambda) \hat{i}+(2-\lambda) \hat{j}+(3+\lambda) \hat{\mathrm{k}})=4+5 \lambda\ \ldots (1)\)
Since this plane is perpendicular to the plane
\(\overrightarrow{\mathrm{r}} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8=0\)
\(\overrightarrow{\mathrm{r}} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})=-8\)
\(-\overrightarrow{\mathrm{r}} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})=8\)
\(\overrightarrow{\mathrm{r}} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})=8 \ldots (2)\)
So, the normal vector of the plane (1) will be perpendicular to the normal vector of plane (2).
Direction ratios of Normal of plane \( (1)=\left(a_{1}, b_{1}, c_{1}\right) \)
\(\equiv[(1-2 \lambda),(2-\lambda),(3+\lambda)]\)
Direction ratios of Normal of plane \( (2)=\left(a_{2}, b_{2}, c_{2}\right) \)
\(\equiv(-5,-3,6)\)
Since the two lines are perpendicular,
\(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)
\(\Rightarrow(1-2 \lambda) \times(-5)+(2-\lambda) \times(-3)+(3+\lambda) \times 6=0\)
\(\Rightarrow-5+10 \lambda-6+3 \lambda+18+6 \lambda=0\)
\(\Rightarrow 19 \lambda+7=0\)
\(\therefore \lambda=-\frac{ 7 }{ 19 }\)
Putting the value of \( \lambda \) in (1), we get -
\(\overrightarrow{\mathrm{r}} \cdot\left(\left(1-2\left(\frac{-7}{19}\right)\right) \hat{i}+\left(2-\left(\frac{-7}{19}\right)\right) \hat{j}+\left(3+\left(\frac{-7}{19}\right)\right) \hat{\mathrm{k}}\right)=4+5\left(\frac{-7}{19}\right)\)
\(\Rightarrow \overrightarrow{\mathrm{r}} \cdot\left(\frac{33}{19} \hat{i}+\frac{45}{19} \hat{j}+\frac{50}{19} \hat{\mathrm{k}}\right)=\frac{41}{19}\)
\(\Rightarrow \overrightarrow{\mathrm{r}} \cdot(33 \hat{i}+45 \hat{j}+50 \hat{\mathrm{k}})=41\)
Hence, the equation of the required plane is
\(\overrightarrow{r} \cdot(33 \hat{i}+45 \hat{j}+50 \hat{k})=41\)

Given -
The equation of line is
\(\overrightarrow{r} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})\)
and the equation of the plane is
\(\overrightarrow{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5\)
To find the intersection of line and plane, putting value of \( \overrightarrow{r} \) from equation of line into equation of plane, we get -
\(\Rightarrow[(2 \hat{i}-\hat{j}+2 \hat{\mathrm{k}})+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{\mathrm{k}})] \cdot(\hat{i}-\hat{j}+\hat{\mathrm{k}})=5\)
\(\Rightarrow[(2+3 \lambda) \hat{i}+(-1+4 \lambda) \hat{j}+(2+2 \lambda) \hat{\mathrm{k}}] \cdot(\hat{i}-\hat{j}+\hat{\mathrm{k}})=5\)
\(\Rightarrow(2+3 \lambda) \times 1+(-1+4 \lambda) \times(-1)+(2+2 \lambda) \times 1=5\)
\(\Rightarrow 2+3 \lambda+1-4 \lambda+2+2 \lambda=5\)
\(\Rightarrow \lambda=0\)
So, the equation of line is
\(\overrightarrow{r}=(2 \hat{i}-\hat{j}+2 \hat{k})\)
Let the point of intersection be \( (x, y, z) \)
So, \( \overrightarrow{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{j}+z \hat{\mathrm{k}} \)
\(\therefore x \hat{i}+y \hat{j}+z \hat{\mathrm{k}}=2 \hat{i}-\hat{j}+2 \hat{\mathrm{k}}\)
Hence, \( x=2, y=-1, z=2 \)
Therefore, the point of intersection is \( (2,-1,2) \).
Now, the distance between points \( \left(x_{1}, y_{1}, z_{1}\right) \) and \( \left(x_{2}, y_{2}, z_{2}\right) \) is given by- \( \sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(z_{1}-z_{2}\right)^{2}} \) units
Distance between the points A \( (2,-1,2) \) and \( B(-1,-5,-10) \) is given by-
\(\mathrm{AB}=\sqrt{(2-(-1))^{2}+(-1-(-5))^{2}+(2-(-10))^{2}}\)
\(=\sqrt{(3)^{2}+(4)^{2}+(12)^{2}}\)
\(=\sqrt{9+16+144}\)
\(=\sqrt{169}\)
\(=13 \text { units }\)

The vector equation of a line passing through a point with position vector \( \overrightarrow{a} \) and parallel to a vector \( \overrightarrow{b} \) is
\(\Rightarrow \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})\)
Given, the line passes through \( (1,2,3) \)
So, \( \overrightarrow{a}=1 \hat{i}+2 \hat{j}+3 \hat{k} \)
Given, line is parallel to both planes
\( \therefore \) Line is perpendicular to normal of both planes.
i.e. \( \overrightarrow{b} \) is perpendicular to normal of both planes.
we know that -
\( \overrightarrow{a} \times \overrightarrow{b} \) is perpendicular to both \( \overrightarrow{a} \& \overrightarrow{b} \)
So, \( \overrightarrow{b} \) is cross product of normal of planes
\( \overrightarrow{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=5 \) and \( \overrightarrow{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=6 \)
\(\begin{array}{l}
\text { Required Normal }=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 2 \\
3 & 1 & 1
\end{array}\right| \end{array}\)
\(=\hat{i}[(-1)(1)-1(3)]-\hat{j}[1(1)-3(2)]+\hat{k}[1(1)-3(-1)] \)
\(=\hat{i}[-1-2]-\hat{j}[1-6]+\hat{\mathrm{k}}[1+3] \)
\(=-3 \hat{i}+5 \hat{j}+4 \hat{\mathrm{k}}\)
Thus, \( \overrightarrow{b}=-3 \hat{i}+5 \hat{j}+4 \widehat{k} \)
Now, putting the value of \( \overrightarrow{a} \& \overrightarrow{b} \) in formula
\(\begin{array}{l}
\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}} \\
=(1 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-3 \hat{i}+5 \hat{j}+4 \hat{k})
\end{array}\)
Therefore, the equation of the line is
\(\overrightarrow{\mathrm{r}}=(1 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-3 \hat{i}+5 \hat{j}+4 \hat{k})\)

The vector equation of a line passing through a point with position vector \( \overrightarrow{a} \) and parallel to a vector \( \overrightarrow{b} \) is
\(\Rightarrow \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\)
Given, the line passes through \( (1,2,-4) \)
So, \( \overrightarrow{a}=1 \hat{i}+2 \hat{j}-4 \hat{k} \)
Given, line is parallel to both planes
\( \therefore \) Line is perpendicular to normal of both planes.
i.e. \( \overrightarrow{b} \) is perpendicular to normal of both planes.
we know that -
\( \overrightarrow{a} \times \overrightarrow{b} \) is perpendicular to both \( \overrightarrow{a} \& \overrightarrow{b} \)
So, \( \overrightarrow{b} \) is cross product of normal of planes
\(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} \text { and } \frac{x-15}{3}=\frac{y+29}{8}=\frac{z-5}{-5} \)
\(\begin{array}{l}\text { Required Normal }=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{\mathrm{k}} \\
3 & -16 & 7 \\
3 & 8 & -5
\end{array}\right|\end{array}\)
\(=\hat{i}[(-16)(-5)-8(7)]-\hat{j}[3(-5)-3(7)]+\hat{\mathrm{k}}[3(8)-3(-16)]\)
\(=\hat{i}[80-56]-\hat{j}[-15-21]+\hat{\mathrm{k}}[24+48]\)
\(=24 \hat{i}+36 \hat{j}+72 \hat{\mathrm{k}}\)
Thus, \( \overrightarrow{b}=24 \hat{i}+36 \hat{j}+72 \hat{k} \)
Now, putting the value of \( \overrightarrow{a} \& \overrightarrow{b} \) in formula
\(\begin{array}{l}
\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}} \\
=(1 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(24 \hat{i}+36 \hat{j}+72 \hat{k}) \\
=(1 \hat{i}+2 \hat{j}-4 \hat{k})+12 \lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\
=(1 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})
\end{array}\)
Therefore, the equation of the line is
\(\overrightarrow{\mathrm{r}}=(1 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\)

Distance of the point \( \left(x_{1}, y_{1}, z_{1}\right) \) from the plane \( \mathrm{Ax}+\mathrm{By}+\mathrm{Cz}=\mathrm{D} \) is
\(\Rightarrow\left|\frac{\mathrm{Ax}_{1}+\mathrm{By}_{1} \mathrm{Cz}_{1}-\mathrm{D}}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{C}^{2}}}\right|\)
The equation of a plane having intercepts \( \mathrm{a}, \mathrm{b}, \mathrm{c} \) on the \( x-, y \)-, \( z \) - axis respectively is
\(\Rightarrow \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
Comparing with \( \mathrm{Ax}+\mathrm{By}+\mathrm{Cz}=\mathrm{D} \), we get -
\(A=\frac{ 1 }{a} ,B= \frac{1}{b}, C=\frac{1 }{c}, D=1\)
Given, the plane is at a distance of ' p ' units from the origin.
So, the point is \( O(0,0,0) \)
\(\therefore x_{1}=0, y_{1}=0, z_{1}=0\)
Now,
Distance \( =\left|\frac{\mathrm{Ax}_{1}+\mathrm{By}_{1}+\mathrm{Cz}_{1}-\mathrm{D}}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{C}^{2}}}\right| \)
Substituting all values, we get -
\(\Rightarrow \mathrm{p}=\left|\frac{\frac{1}{a} \times 0+\frac{1}{b} \times 0+\frac{1}{c} \times 0-1}{\sqrt{\left(\frac{1}{a}\right)^{2}+\left(\frac{1}{b}\right)^{2}+\left(\frac{1}{c}\right)^{2}}}\right|\)
\(\Rightarrow \mathrm{p}=\left|\frac{0+0+0-1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}}\right|\)
\(\Rightarrow \mathrm{p}=\left|\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}}\right|\)
\(\Rightarrow \frac{1}{\mathrm{p}}=\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}\)
squaring both sides, we get -
\(\Rightarrow \frac{1}{\mathrm{p}^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\)
Hence Proved.

Distance between two parallel planes \( \mathrm{Ax}+\mathrm{B} y+\mathrm{Cz}=\mathrm{d}_{1} \) and \( \mathrm{Ax}+\mathrm{B} y+ \) \( \mathrm{C} z=\mathrm{d}_{2} \) is
\(\Rightarrow\left|\frac{\mathrm{d}_{1}-\mathrm{d}_{2}}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{C}^{2}}}\right|\)
Given -
First Plane is
\(2 x+3 y+4 z=4\)
Comparing with \( \mathrm{Ax}+\mathrm{B} y+\mathrm{C} z=\mathrm{d}_{1} \), we get -
\(A=2, B=3, C=4, d_{1}=4\)
Second Plane is
\(4 x+6 y+8 z=12\)
After Dividing by 2,
\(2 x+3 y+4 z=6\)
Comparing with \( \mathrm{A} x+\mathrm{B} y+\mathrm{Cz}=\mathrm{d}_{1} \), we get -
\(A=2, B=3, C=4, d_{2}=6\)
So,
Distance between two planes
\(=\left|\frac{4-6}{\sqrt{2^{2}+3^{2}+4^{2}}}\right|\)
\(=\left|\frac{-2}{\sqrt{4+9+16}}\right|\)
\(= \frac{ 2 }{ \sqrt{29} }\)
Hence, (D) is the correct option.

Given -
First Plane is
\(2 x-y+4 z=5\)
Multiply both sides by 2.5 , we get -
\(5 x-2.5 y+10 z=12.5\ldots (1)\)
Second Plane is
\(5 x-2.5 y+10 z=6 \ldots (2)\)
Clearly, the direction ratios of normal of both the plane (1) and (2) are same.
Hence, Both the given planes are parallel.