Ex 11.3 Class 12 Maths Ncert Solutions

The eq. of the plane
\(\mathrm{z}=2\)
Direction ratio of the normal \( (0,0,1) \)
\(\therefore \sqrt{(0)^{2}+(0)^{2}+(1)^{2}}\)
\(=\sqrt{1}\)
\(\frac{0}{1} x+\frac{0}{1} y+\frac{1}{1} z=\frac{2}{1}\)
This is the form of
\( l x+m y+n z=\mathrm{d}(\therefore \mathrm{d}= \) Distance of the normal from the origin. \( ) \)
Direction cosines \( =0,0,1 \)
Distance \( (d)=2 \)

The eq. of the plane
\(x+y+z=1\)
Direction ratio of the normal \( (1,1,1) \)
\(\therefore \sqrt{(1)^{2}+(1)^{2}+(1)^{2}}\)
\(=\sqrt{3}\)
\(\frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}} y+\frac{1}{\sqrt{3}} z=\frac{1}{\sqrt{3}}\)
This is the form of
\( l x+m y+n z=d(\therefore \mathrm{d}= \) Distance of the normal from the origin. \( ) \)
Direction cosines \( =\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \)
Distance \( (d)=\frac{1}{\sqrt{3}} \)

The eq. of the plane
\(2 x+3 y-z=5\)
Direction ratio of the normal \( (2,3,-1) \)
\( \therefore \sqrt{(2)^{2}+(3)^{2}+(-1)^{2}} \)
\( =\sqrt{14} \)
\( \frac{2}{\sqrt{14}} x+\frac{3}{\sqrt{14}} y+\frac{1}{\sqrt{14}} z=\frac{5}{\sqrt{14}} \)
This is the form of
\( l x+m y+n z=\mathrm{d}(\therefore \mathrm{d}= \) Distance of the normal from the origin. \( ) \)
Direction cosines \( =\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}} \)
Distance \( (d)=\frac{5}{\sqrt{14}} \)

The eq. of the plane
\(0 x-5 y+0 z=8\)
Direction ratio of the normal \( (0,-5,0) \)
\(\therefore \sqrt{(0)^{2}+(-5)^{2}+(0)^{2}}\)
\(=\sqrt{25}\)
\(=5\)
\(\frac{0}{5} x-\frac{5}{5} y-\frac{0}{5} z=\frac{8}{5}\)
This is the form of
\( l x+m y+n z=\mathrm{d}(\therefore \mathrm{d}= \) Distance of the normal from the origin. \( ) \)
Direction cosine \( =0,-1,0 \)
Distance \( (\mathrm{d})=\frac{8}{5} \)

Vector eq. of the plane with position vector \( \overrightarrow{r} \) is
\( \overrightarrow{\mathrm{r}} \hat{n}=\mathrm{d} \)
\( \hat{\mathrm{n}}=\frac{\overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}|}=\frac{3 \hat{i}+5 \hat{j}-6 \hat{k}}{\sqrt{9+25+36}} \)
\( =\frac{3 \hat{i}+5 \hat{j}-6 \widehat{k}}{\sqrt{70}} \)
\( \overrightarrow{\mathrm{r}} \). \( \frac{3 \hat{\mathrm{i}}+5 \hat{j}-6 \widehat{\mathrm{k}}}{\sqrt{70}}=7 \)
\( \overrightarrow{\mathrm{r}} \cdot 3 \hat{i}+5 \hat{j}-6 \hat{k}=7 \sqrt{70} \)

Let \( \overrightarrow{r} \) be the position vector of \( \mathrm{P}(x, y, z) \)
Hence, \( \overrightarrow{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{j}+z \hat{\mathrm{k}} \)
\( \therefore \overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{j}-\hat{\mathrm{k}})=2 \)
\( (x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}-\hat{\mathrm{k}})=2 \)
So, Cartesian eq. is
\(x+y-z=2\)

Let \( \overrightarrow{\mathrm{r}} \) be the position vector of \( \mathrm{P}(x, y, z) \)
Hence, \( \overrightarrow{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{j}+z \hat{\mathrm{k}} \)
\(\therefore \overrightarrow{\mathrm{r}} \cdot({2} \hat{i}+{3} \hat{j}-{4} \hat{k})=1\)
\((x \hat{i}+y \hat{j}+z \hat{\mathrm{k}}) \cdot(\widehat{2 \mathrm{i}}+\widehat{3 \mathrm{j}}-\widehat{4 \mathrm{k}})=1\)
So, Cartesian eq. is
\(2 x+3 y-4 z=1\)

Let \( \overrightarrow{\mathrm{r}} \) be the position vector of \( \mathrm{P}(x, y, z) \)
Hence, \( \overrightarrow{\mathrm{r}}=x \hat{i}+y \hat{j}+z \hat{\mathrm{k}} \)
\(\therefore \overrightarrow{\mathrm{r}} .[(\mathrm{s}-2 \mathrm{t}) \hat{i}+(3-\mathrm{t}) \hat{j}+(2 \mathrm{~s}+\mathrm{t}) \hat{\mathrm{k}}]=15\)
\((x \hat{i}+y \hat{j}+z \hat{\mathrm{k}}) \cdot[(\mathrm{s}-2 \mathrm{t}) \hat{i}+(3-\mathrm{t}) \hat{j}+(2 \mathrm{~s}+\mathrm{t}) \hat{\mathrm{k}}]=15\)
So, Cartesian eq. is
\((s-2 t) \hat{i}+(3-t) \hat{j}+(2 s+t) \hat{k}=15\)

Let the coordinate of the foot of \( \perp \mathrm{P} \) from the origin to the given plane be P \( (x, y, z) \).
\(2 x+3 y+4 z=12\)
Direction ratio \( (2,3,4) \)
\(\begin{array}{l}
\therefore \sqrt{(2)^{2}+(3)^{2}+(4)^{2}} \\
=\sqrt{4+9+16} \\
=\sqrt{29} \\
\frac{2}{\sqrt{29}} x+\frac{3}{\sqrt{29}} y+\frac{4}{\sqrt{29}} z=\frac{12}{\sqrt{29}}
\end{array}\)
This is the form of
\( l x+m y+n z=\mathrm{d}(\therefore \mathrm{d}= \) Distance of the normal from the origin.
Direction cosines \( =\frac{2}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}} \)
Coordinate of the foot \( (l d, m d, n d)=\frac{24}{29}, \frac{36}{9}, \frac{48}{29} \)

Let the coordinate of the foot of \( \perp \mathrm{P} \) from the origin to the given plane be \( \mathrm{P}(x, y, z) \).
\(0 x+3 y+4 z=6\)
Direction ratio \( (0,3,4) \)
\(\therefore \sqrt{(0)^{2}+(3)^{2}+(4)^{2}}\)
\(=\sqrt{0+9+16}\)
\(=\sqrt{25}=5\)
\(\frac{0}{5} x+\frac{3}{5} y+\frac{4}{5} z=\frac{6}{5}\)
This is the form of
\( l x+m y+n z=\mathrm{d}(\therefore \mathrm{d}= \) Distance of the normal from the origin.
Direction cosines \( =\frac{0}{5}, \frac{3}{5}, \frac{4}{5} \)
Coordinate of the foot \( (l d, m d, n d)=0, \frac{18}{25}, \frac{24}{25} \)

Let the coordinate of the foot of \( \perp \mathrm{P} \) from the origin to the given plane be \( \mathrm{P}(x, y, z) \).
\(x+y+z=1\)
Direction ratio \( (0,3,4) \)
\( \therefore \sqrt{(1)^{2}+(1)^{2}+(1)^{2}} \)
\( =\sqrt{1+1+1} \)
\( =\sqrt{3} \)
\( \frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}} y+\frac{1}{\sqrt{3}} z=\frac{1}{\sqrt{3}} \)
This is the form of
\( l x+m y+n z=\mathrm{d}(\therefore \mathrm{d}= \) Distance of the normal from the origin. \( ) \)
Direction cosines \( =\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \)
Coordinate of the foot \( (l d, m d, n d)=\frac{1}{3}, \frac{1}{3}, \frac{1}{3} \)

Let the coordinate of the foot of \( \perp \mathrm{P} \) from the origin to the given plane be \( \mathrm{P}(x, y, z) \).
\(x-5 y+0 z=1\)
Direction ratio \( (0,-5,0) \)
\(\therefore \sqrt{(0)^{2}+(-5)^{2}+(0)^{2}}\)
\(=\sqrt{0+25+0}\)
\(=\sqrt{25}\)
\(=5\)
\(\frac{0}{5} x+\frac{5}{5} y+\frac{0}{5} z=\frac{8}{5}\)
This is the form of
\( l x+m y+n z=\mathrm{d}(\therefore \mathrm{d}= \) Distance of the normal from the origin. \( ) \)
Direction cosines \( =0,-1,0 \)
Coordinate of the foot \( (l d, m d, n d)=0,-\frac{8}{5}, 0 \)

Let the position vector of the point
\(\overrightarrow{a}=(1 \hat{i}+0 \hat{j}-2 \hat{k})\)
Normal \( \overrightarrow{\mathrm{N}} \perp \) to the plane
\(\overrightarrow{N}=\hat{i}+\hat{j}-\hat{k}\)
Vector eq. of the plane,
\((\overrightarrow{r}-\overrightarrow{a}) \cdot \overrightarrow{N}=0\)
\((\overrightarrow{r}-(\hat{i}-2 \hat{k})) \cdot \hat{i}+\hat{j}-\hat{k}=0\)
\(\text { As, } \overrightarrow{\mathrm{r}}=x \hat{i}+y \hat{j}+z \hat{\mathrm{k}}\)
\(\therefore(x \hat{i}+y \hat{j}+z \hat{\mathrm{k}}-\hat{i}+2 \hat{\mathrm{k}}) \cdot \hat{i}+\hat{j}-\hat{\mathrm{k}}=0\)
\(x-1+y-z-2=0\)
\(x+y-z-3=0\)
Required Cartesian eq. of the plane
\(x+y-z=3\)

Let the position vector of the point
\(\overrightarrow{a}=(1 \hat{i}+4 \hat{j}+6 \hat{k})\)
Normal \( \overrightarrow{\mathrm{N}} \perp \) to the plane
\(\overrightarrow{N}=\hat{i}-2 \hat{j}+\hat{k}\)
Vector eq. of the plane,
\((\overrightarrow{r}-\overrightarrow{a}) \cdot \overrightarrow{N}=0\)
\((\overrightarrow{r}-(1 \hat{i}+4 \hat{j}+6 \hat{k})) \cdot \hat{i}-2 \hat{j}+\hat{k}=0\)
As, \( \overrightarrow{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{j}+z \hat{\mathrm{k}} \)
\(\begin{array}{l}
\therefore(x \hat{i}+y \hat{j}+z \hat{\mathrm{k}}-\hat{i}-4 \hat{j}+6 \hat{\mathrm{k}}) \cdot \hat{i}-2 \hat{j}+\hat{\mathrm{k}}=0 \\
{[(x-1) \hat{i}+(y-4) \hat{j}+(z-6) \hat{\mathrm{k}}] \cdot \hat{i}-2 \hat{j}+\hat{\mathrm{k}}=0} \\
x-1-2 y+8+z-6=0 \\
x-2 y+z+1=0
\end{array}\)
Required Cartesian eq. of the plane
\(x-2 y+z=-1\)

The given points are \( (1,1,-1),(6,4,-5),(-4,-2,3) \).
Let,
\(\begin{array}{l}
=\left|\begin{array}{lrr}
1 & 1 & -1 \\
6 & 4 & -5 \\
-4 & -2 & 3
\end{array}\right| \\
=1(12-10)-1(18-20)-1(-12+16) \\
=2+2-4=0
\end{array}\)
Since, the value of determinant is 0 .
Therefore, these points are collinear as there will be infinite planes passing through the given 3 points.

The given points are \( (1,1,0),(1,2,1),(-2,2,-1) \)
Let,
\(\begin{array}{l}
=\left|\begin{array}{lrr}
1 & 1 & 0 \\
1 & 2 & 1 \\
-2 & 2 & -1
\end{array}\right| \\
=1(-2-2)-1(-1+2) \\
=-4-1 \\
=-5 \neq 0
\end{array}\)
There passes a unique plane from the given 3 points.
Equation of the plane passes through the points, \( \left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right) \) and \( \left(x_{3}, y_{3}, z_{3}\right) \), i.e.,
\(\begin{array}{l}
=\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right| \\
=\left|\begin{array}{ccc}
x-1 & y-1 & z \\
x_{2}-1 & y_{2}-1 & z_{2} \\
x_{3}-1 & y_{3}-1 & z_{3}
\end{array}\right|
\end{array}\)
\(=\left|\begin{array}{ccc}
x-1 & y-1 & z \\
1-1 & 2-1 & 1 \\
-2-1 & 2-1 & -1
\end{array}\right|\)
\(\begin{array}{l}
=\left|\begin{array}{ccc}
x-1 & y-1 & z \\
0 & 1 & 1 \\
-3 & 1 & -1
\end{array}\right| \\
\Rightarrow(x-1)(-2)-(y-1)(3)+3 z=0 \\
\Rightarrow-2 x+2-3 y+3+3 z=0 \\
\Rightarrow 2 x+3 y-3 z=5
\end{array}\)
This is the required eq. of the plane.

We know that, the eq. of the plane in intercept form
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
where \( \mathrm{a}, \mathrm{b}, \mathrm{c} \) is the intercepts cut-off by the plane at \( \mathrm{x}, \mathrm{y} \) and z axes respectively.
\(\Rightarrow 2 x+y-z=5\ldots(i)\)
Dividing both side of (i)eq. by 5 , we get
\(\Rightarrow \frac{2}{5} x+\frac{y}{5}-\frac{z}{5}=\frac{5}{5}\)
\(\Rightarrow \frac{2}{5} x+\frac{y}{5}-\frac{z}{5}=1\)
\(\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}=1\)
Here, \( \mathrm{a}=\frac{5}{2}, \mathrm{~b}=5 \) and \( \mathrm{c}=-5 \)
Thus, the intercepts cut-off by the plane are \( \frac{5}{2},5 \) and -5 .

We know that the eq. of the plane \(ZOX\) is
\(y=0\)
Eq. of plane parallel to it is of the form, \( y=\mathrm{a} \)
Hence, the required eq. of the plane is
\(y=3\)

Eq. of the plane passes through the intersection of the plane is given by
\((3 x-y+2 z-4)+\lambda(x+y+z-2)=0\)
\( \because \) Plane passes through the points \( (2,2,1) \)
\((3 \times 2-2+2 \times 1-4)+\lambda(2+2+1-2)=0\)
\(2+3 \lambda=0\)
\(3 \lambda=-2\)
\(\lambda=\frac{-2}{3} \ldots\text { (i) }\)
Hence, the required eq. of the plane
\((3 x-y+2 z-4)-\frac{2}{3}(x+y+z-2)=0\)
\(\frac{9 x-3 y+6 z-12-2 x-2 y-2 z+4}{3}=0\)
\(7 x-5 y+4 z-8=0\)
This is the required eq. of the plane.

Let the vector eq. of the plane passing through the intersection of the planes
\( \overrightarrow{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7 \), and \(\overrightarrow{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9 \)
Here, \( \overrightarrow{r} .(2 \hat{i}+2 \hat{j}-3 \hat{k})-7=0 \)
\( \overrightarrow{r} .(2 \hat{i}+5 \hat{j}+3 \hat{k})-9=0 \)
\( \therefore[\overrightarrow{\mathrm{r}} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{\mathrm{k}})-7]+\lambda[\overrightarrow{\mathrm{r}} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{\mathrm{k}})-9]=0 \)
\( \overrightarrow{\mathrm{r}} \cdot[(2 \hat{i}+2 \hat{j}-3 \hat{\mathrm{k}})+(2 \lambda \hat{i}+5 \lambda \hat{j}+3 \lambda \hat{\mathrm{k}})]-7-9 \lambda=0 \)
\( \overrightarrow{\mathrm{r}} \cdot[(2+2 \lambda) \hat{i}+(2+5 \lambda) \hat{j}+(-3+3 \lambda) \hat{\mathrm{k}}]-7-9 \lambda=0 \)
\( \because \) Plane passes through points \( (2,1,3) \)
\((2 \hat{i}+\hat{j}+3 \hat{k}) \cdot[(2+2 \lambda) \hat{i}+(2+5 \lambda) \hat{j}+(-3+3 \lambda) \hat{k}]-7-9 \lambda=0\)
\(4+4 \lambda+2+5 \lambda-9+9 \lambda-7-9 \lambda=0\)
\(9 \lambda=10\)
\(\lambda=\frac{10}{9}\)
\(\overrightarrow{\mathrm{r}} \cdot \left[\left(2+\frac{20}{9}\right) \hat{i}+\left(2+\frac{50}{9}\right) \hat{j}+\left(-3+\frac{30}{9}\right) \hat{\mathrm{k}}\right]-7-9 \frac{10}{9}=0\)
\(\overrightarrow{\mathrm{r}} .\left[\left(2+\frac{20}{9}\right) \hat{i}+\left(2+\frac{50}{9}\right) \hat{j}+\left(-3+\frac{30}{9}\right) \hat{k}\right]-17=0\)
\(\overrightarrow{\mathrm{r}} \cdot\left[\left(2+\frac{20}{9}\right) \hat{i}+\left(2+\frac{50}{9}\right) \hat{j}+\left(-3+\frac{30}{9}\right) \hat{\mathrm{k}}\right]=17\)
\(\overrightarrow{\mathrm{r}} \cdot\left[\frac{38}{9} \hat{i}+\frac{68}{9} \hat{j}+\frac{3}{9} \hat{k}\right]=17\)
\(\overrightarrow{\mathrm{r}} \cdot[38 \hat{i}+68 \hat{j}+3 \hat{k}]=153\)
This is the required vector eq. of the plane.

Let the eq. of the plane that passes through the two-given plane \( x+y+z \) \( =1 \) and \( 2 x+3 y+4 z=5 \) is
\((x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0\)
\((2 \lambda+1) x+(3 \lambda+1) y+(4 \lambda+1) z-1-5 \lambda=0\) (i)
Direction ratio of the plane \( (2 \lambda+1,3 \lambda+1,4 \lambda+1) \)
and Direction ratio of another plane \( (1,-1,1) \)
\( \because \) Both are \( \perp \) hence
\((2 \lambda+1 \times 1)+(3 \lambda+1 \times(-1))+(4 \lambda+1 \times 1)=0\)
\(2 \lambda+1-3 \lambda-1+4 \lambda+1=0\)
\(\lambda=\frac{-1}{3}\)
Put the value of \( \lambda \) in (i)eq., we get
\(\left(2 \frac{(-1)}{3}+1\right) x+\left(3 \frac{(-1)}{3}+1\right) y+\left(4 \frac{(-1)}{3}+1\right) z\)
\(\frac{1}{3} x-\frac{1}{3} z+\frac{2}{3}=0\)
\(x-z+2=0\)
This is the required eq. of the plane.

The eq. of the given planes are
\( \overrightarrow{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5 \), and \( \overrightarrow{r} \cdot(3 \hat{i}-3 \hat{j}+5 \hat{k})=5 \)
If \( n_{1} \) and \( n_{2} \) are normal to the planes, \( \overrightarrow{r_{1}} \cdot \overrightarrow{n_{1}}=d_{2} \) and \( \overrightarrow{r_{2}} \cdot \overrightarrow{n_{2}}=d_{2} \)
Angle between two planes
\(\cos \theta=\left|\frac{\overrightarrow{\mathrm{n}_{1}} \cdot \overrightarrow{\mathrm{n}_{2}}}{\left|\overrightarrow{\mathrm{n}_{1}} \cdot \overrightarrow{\mathrm{n}_{2}}\right|}\right|\)
\(=\left|\frac{6-6-15}{\sqrt{4+4+9} \sqrt{9+9+25}}\right|\)
\(=\left|\frac{-15}{\sqrt{17} \sqrt{43}}\right|\)
\(\theta=\cos ^{-1}\left(\frac{-15}{\sqrt{17} \sqrt{43}}\right)\)
\(=\cos ^{-1}\left(\frac{15}{\sqrt{731}}\right)\)

The eq. of the given planes are
\(7 x+5 y+6 z+30=0 \text { and } 3 x-y-10 z+4=0\)
Two planes are \( \perp \) if the direction ratio of the normal to the plane is
\(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)
\(21-5-60\)
\(-44 \neq 0\)
\( \therefore \) Both the planes are not \( \perp \) to each other.
Two planes are \(||\) to each other if the direction ratio of the normal to the plane is
\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{~b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\)
\(\frac{7}{3} \neq \frac{5}{-1} \neq \frac{6}{-10}\)
\( \therefore \) Both the planes are not \( \| \) to each other.
The angle between them is given by
\(\begin{array}{l}
\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right| \\
=\left|\frac{-44}{\sqrt{49+25+36} \sqrt{9+1+100}}\right| \\
=\left|\frac{-44}{\sqrt{110} \sqrt{110}}\right|=\frac{-44}{110} \\
\theta=\cos ^{-1} \frac{2}{5}
\end{array}\)

The eq. of the given planes is are
\(2 x+y+3 z-2=0 \text { and } x-2 y+5=0\)
Two planes are \( \perp \) if the direction ratio of the normal to the plane is
\(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)
\(2 \times 1+1 \times(-2)+3 \times 0\)
\(=0\)
Thus, the given planes are \( \perp \) to each other.

The eq. of the given planes are
\(2 x-2 y+4 z+5=0 \text { and } x-2 y+5=0\)
Two planes are \( \perp \) if the direction ratio of the normal to the plane is
\(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)
\(6+6+24\)
\(36 \neq 0\)
\( \therefore \) Both the planes are not \( \perp \) to each other.
Two planes are \(\|\) to each other if the direction ratio of the normal to the plane is
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
\(\frac{2}{3}=\frac{-2}{-3}=\frac{4}{6}\)
\(\frac{2}{3}=\frac{2}{3}=\frac{2}{3}\)
Thus, the given planes are \( \| \) to each other.

The eq. of the given planes are
\(2 x-y+3 z-1=0 \text { and } 2 x-y+3 z+3=0\)
Two planes are \( \perp \) if the direction ratio of the normal to the plane is
\(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)
\(2 \times 2+(-1) \times(-1)+3 \times 3\)
\(14 \neq 0\)
\( \therefore \) Both the planes are not \( \perp \) to each other.
Two planes are \( \| \) to each other if the direction ratio of the normal to the plane is
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
\(\frac{2}{2}=\frac{-1}{-1}=\frac{3}{5}\)
\(\frac{1}{1}=\frac{1}{1}=\frac{1}{1}\)
Thus, the given planes are \( \| \) to each other.

The eq. of the given planes are
\(4 x+8 y+z-8=0 \text { and } y+z-4=0\)
Two planes are \( \perp \) if the direction ratio of the normal to the plane is
\(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)
\(0+8+1\)
\(9 \neq 0\)
\( \therefore \) Both the planes are not \( \perp \) to each other.
Two planes are \(\|\) to each other if the direction ratio of the normal to the plane is
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
\(\frac{4}{0} \neq \frac{8}{1} \neq \frac{1}{1}\)
\( \therefore \) Both the planes are not \( \| \) to each other.
The angle between them is given by
\(\begin{array}{l}
\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right| \\
=\left|\frac{4 \times 0+8 \times 1+1 \times 1}{\sqrt{16+64+1} \sqrt{0+1+1}}\right|
\end{array}\)
\(\begin{array}{l}
=\frac{9}{9 \sqrt{2}} \\
\theta=\cos ^{-1} \frac{9}{9 \sqrt{2}} \\
=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}
\end{array}\)

Distance of point \( \mathrm{P}\left(x_{1}, y_{1}, z_{1}\right) \) from the plane \( \mathrm{Ax}+\mathrm{By}+\mathrm{Cz}-\mathrm{D}=0 \) is
\(\mathrm{d}=\left|\frac{\mathrm{Ax}_{1}+\mathrm{By}_{1}+\mathrm{Cz}_{1}-\mathrm{D}}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{C}^{2}}}\right|\)
Given point is \( (0,0,0) \) and the plane is \( 3 x-4 y+12 z=3 \)
\(d=\left|\frac{0+0+0+3}{\sqrt{9+16+144}}\right|\)
\(=\left|\frac{3}{\sqrt{169}}\right|=\frac{3}{13}\)

Given point is \( (3,-2,1) \) and the plane is \( 2 x-y+2 z+3=0 \)
\(d=\left|\frac{6+2+2+3}{\sqrt{4+1+4}}\right|\)
\(=\left|\frac{13}{\sqrt{9}}\right|=\frac{13}{3}\)

Given point is \( (2,3,-5) \) and the plane is \( x+2 y-2 z=9 \)
\(d=\left|\frac{2+6+10-9}{\sqrt{1+4+4}}\right|\)
\(=\left|\frac{9}{\sqrt{9}}\right|\)
\(=\frac{9}{3}=3\)

Given point is \( (-6,0,0) \) and the plane is \( 2 x-3 y+6 z-2=0 \)
\(d=\left|\frac{-12-0+0-2}{\sqrt{4+9+36}}\right|\)
\(=\left|\frac{14}{\sqrt{49}}\right|\)
\(=\frac{14}{7}=2\)