RD Sharma Class 12 Exercise 28.11 The Plane Solutions Maths - Download PDF Free Online

The Plane Excercise: 28.11

The Plane Exercise 28.11 Question 1

Answer:${\sin }^{-1}\frac{9}{\sqrt{89}}$
Hint: Use formula $\overrightarrow{r}=(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+9\hat{k})+\lambda (2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{k})\text{ and }\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})=5$
Given:
Solution: Equation of line is $\overrightarrow{r}=(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+9\hat{k})+\lambda (2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{k})$ and the equation of plane is $\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})=5$
As we know that the angle $\theta$ between the line $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and plane $a_{2}x+b_{2}y+c_{2}z+d_2=0$
$\sin \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
Here, $a_1=1,b_1=1,c_1=1$
and $a_2=2,b_2=3,c_2=4$
The angle between them is
$\sin \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\begin{array}{rl}& =\frac{1\times 2+1\times 3+1\times 4}{\sqrt{1^2+1^2+1^2}\sqrt{2^2+3^2+4^2}}\\ & \end{array}$
$=\frac{2+3+4}{\sqrt{3}\sqrt{29}}$
$=\frac{9}{\sqrt{87}}$
$\theta ={\sin }^{-1}\frac{9}{\sqrt{87}}$

The Plane Exercise 28.11 Question 2

Answer:$0$
Hint: Use formula $\sin \theta =\frac{\overrightarrow{b}\cdot \overrightarrow{n}}{|\overrightarrow{b}||\overrightarrow{n}|}$
Given: Line is $\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{1}$ and plane is $2x+y-z=4$
Solution: The given line is parallel to the vector $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and the given plane is normal to the vector $\overrightarrow{n}=2\hat{i}+\hat{j}-\hat{k}$
We know that, the angle $\theta$ between the line and plane is
$\begin{array}{rl}& \sin \theta =\frac{(\hat{i}-\hat{j}+\hat{k})\cdot (2\hat{i}+\hat{j}-\hat{k})}{|\hat{i}-\hat{j}+\hat{k}||2\hat{i}+\hat{j}-\hat{k}|}\\ & \end{array}$ $[\sin \theta =\frac{\overrightarrow{b}\cdot \overrightarrow{n}}{|\overrightarrow{b}||\overrightarrow{n}|}]$
$=\frac{2-1-1}{\sqrt{1+1+1}\sqrt{4+1+1}}=0$
$\Rightarrow \theta ={\sin }^{-1}0$
$\theta =0$

The Plane Exercise 28.11 Question 3

Answer:${\sin }^{-1}\left(\frac{23}{11\sqrt{11}}\right)$
Hint: Use formula $\sin \theta =\frac{\overrightarrow{b}\cdot \overrightarrow{n}}{|\overrightarrow{b}||\overrightarrow{n}|}$
Given:$(3,-4,-2)\mathrm{\&}(12,2,0)\text{ and plane }3x-y+z=1$
Solution: It is given that the line passes through $(3,-4,-2)\mathrm{\&}(12,2,0)$
So, $\begin{array}{rl}\overrightarrow{b}=\overrightarrow{\mathrm{AB}}& =\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\\ & \end{array}$
$=12\hat{\mathrm{i}}+2\hat{\mathrm{j}}+0\hat{\mathrm{k}}-(3\hat{\mathrm{i}}-4\hat{\mathrm{j}}-2\hat{\mathrm{k}})$
$=9\hat{\mathrm{i}}+6\hat{\mathrm{j}}+2\hat{\mathrm{k}}$
The given line is parallel to the vector $\overrightarrow{\mathrm{b}}=9\hat{\mathrm{i}}+6\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and given line is normal to the vector $\overrightarrow{n}=3\hat{i}-\hat{j}+\hat{k}$
We know that, the angle $\theta$ between the line and plane is given by
$\begin{array}{rl}\sin \theta & =\frac{\overrightarrow{b}\cdot \overrightarrow{n}}{|\overrightarrow{b}||\overrightarrow{n}|}\\ & \end{array}$
$=\frac{(9\hat{i}+6\hat{j}+2\hat{k})\cdot (3\hat{i}-\hat{j}+\hat{k})}{|9\hat{i}+6\hat{j}+2\hat{k}\Vert 3\hat{i}-\hat{j}+\hat{k}|}$
$=\frac{27-6+2}{\sqrt{81+36+4}\sqrt{9+1+1}}=0$
$=\frac{23}{11\sqrt{11}}$
$\Rightarrow \theta ={\sin }^{-1}\frac{23}{11\sqrt{11}}$

The Plane exercise 28.11 question 4

Answer:$m=-3$
Hint: Use properties of vector
Given: Line $\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\lambda (2\hat{\mathrm{i}}-\mathrm{m}\hat{\mathrm{j}}-3\hat{\mathrm{k}})$ and
Plane $\overrightarrow{r}\cdot (m\hat{i}+3\hat{j}+\hat{k})=4$
Solution: The given line is parallel to the vector $\overrightarrow{\mathrm{b}}=2\hat{\mathrm{i}}-\mathrm{m}\hat{\mathrm{j}}-3\hat{\mathrm{k}}$ and the given plane is normal to the vector $\overrightarrow{\mathrm{n}}=m\hat{\mathrm{i}}+3\hat{\mathrm{j}}+\hat{\mathrm{k}}$
If the line is parallel to the plane, the normal to the plane is perpendicular to the line
$\begin{array}{rl}& \Rightarrow \overrightarrow{b}\perp \overrightarrow{n}\\ & \Rightarrow \overrightarrow{b}\cdot \overrightarrow{n}=0\\ & \Rightarrow (2\hat{i}-m\hat{j}-3\hat{k})(m\hat{i}+3\hat{j}+\hat{k})=0\\ & \Rightarrow 2m-3m-3=0\\ & \Rightarrow -m-3=0\\ & \Rightarrow m=-3\end{array}$

The Plane exercise 28.11 question 5

Answer: Therefore, the given line is parallel to given plane is showed. And distance is $\frac{7}{\sqrt{3}}$ unit
Hint: Use properties of vector and formula $d=\frac{|\overrightarrow{a}\cdot \overrightarrow{n}-d|}{|\overrightarrow{n}|}$
Given: $\overrightarrow{\mathrm{r}}=2\hat{\mathrm{i}}+5\hat{\mathrm{j}}+7\hat{\mathrm{k}}+\lambda (\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})\text{ and }\overrightarrow{\mathrm{r}}\cdot (\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=7$
Solution: The given plane passes through the point with position vector $\overrightarrow{a}=2\hat{i}+5\hat{j}+7\hat{k}$ and is parallel to the vector $\overrightarrow{b}=\hat{i}+3\hat{j}+4\hat{k}$
The given plane is $\overrightarrow{r}\cdot (\hat{i}+\hat{j}-\hat{k})=7$
So, the normal vector $\overrightarrow{n}=\hat{i}+\hat{j}-\hat{k}\text{ and }d=7$
Now, $\overrightarrow{b}\cdot \overrightarrow{n}=(\hat{i}+3\hat{j}+4\hat{k})\cdot (\hat{i}+\hat{j}-\hat{k})=1+3-4=0$
So, $\overrightarrow{b}$ is perpendicular to $\overrightarrow{n}$
So, the given line is parallel to the given plane.
The distance between the line and the parallel plane.
Then, d = length of perpendicular from the point $\overrightarrow{\mathrm{a}}=2\hat{\mathrm{i}}+5\hat{\mathrm{j}}+7\hat{\mathrm{k}}$ to the plane $\overrightarrow{r}\cdot \overrightarrow{n}=d$
$\begin{array}{rl}d& =\frac{|\overrightarrow{a}\cdot \overrightarrow{n}-d|}{|\overrightarrow{n}|}\\ & \end{array}$
$=\frac{|(2\hat{i}+5\hat{j}+7\hat{k})(\hat{i}+\hat{j}-\hat{k})-7|}{|\hat{i}+\hat{j}-\hat{k}|}$
$\begin{array}{rl}& =\frac{|2+5-7-7|}{\sqrt{1+1+1}}\\ & \end{array}$
$=\frac{7}{\sqrt{3}}\text{ units }$

The Plane exercise 28.11 question 6

Answer:$\overrightarrow{r}=\lambda (\hat{i}+2\hat{j}+3\hat{k})$
Hint: We use properties of vector
Given:$\overrightarrow{r}\cdot (\hat{i}+2\hat{j}+3\hat{k})=3$
Solution: The requires line is perpendicular to the plane $\overrightarrow{r}\cdot (\hat{i}+2\hat{j}+3\hat{k})=3$
Therefore, it is parallel to the normal $\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}$
Thus, the required line passes through the position vector $\overrightarrow{\mathrm{a}}=0\hat{\mathrm{i}}+0\hat{\mathrm{j}}+0\hat{\mathrm{k}}$ and parallel to the vector $\overrightarrow{\mathrm{n}}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}$
So, its vector equation is
$\begin{array}{rl}& \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{n}\\ & \overrightarrow{r}=(0\hat{i}+0\hat{\mathrm{j}}+0\hat{k})+\lambda (\hat{\mathbf{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}})\\ & \overrightarrow{r}=\lambda (\hat{\mathbf{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}})\end{array}$

The Plane Excercise 28.11 Question 7

Answer:$7y+4z-5=0$
Hint: Use formula $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
Given: $(2,3,-4)\mathrm{\&}(1,-1,3)$
Solution: We know that the equation of plane passing through $({\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1)$ is given by
$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ ……………….. (1)
So, equation of plane passing through (2,3,-4) is
$a(x-2)+b(y-3)+c(z+4)=0$ …………….…… (2)
It also passes through (1,-1,3)
$\Rightarrow -a-4b+7c=0$
$a(1-2)+b(-1-3)+c(3+4)=0$
$\begin{array}{rlr}& & \Rightarrow a+4b-7c=0\end{array}$ ……………….... (3)
We know that line
$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ is parallel to plane $a_{2}x+b_{2}y+c_{2}z+d_2=0$
if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$ ………………….. (4)
Here equation (2) is parallel to x- axis
$\frac{x}{1}=\frac{y}{0}=\frac{x}{0}$ …………………… (5)
Using (2) and (5) in equation (4) we get
$\begin{array}{rl}& a\times 1+b\times 0+c\times 0=0\\ & \Rightarrow a=0\end{array}$
Putting the value of ‘a’ in equation (3) we get
$\begin{array}{rl}& \Rightarrow a-4b+7c=0\\ & \end{array}$
$\Rightarrow 0-4b+7c=0$
$\Rightarrow -4b=-7c$
$\Rightarrow b=\frac{7c}{4}$
Now, putting the value of a and b in (2) we get
$\begin{array}{rl}& a(x-2)+b(y-3)+c(z+4)=0\\ & \end{array}$
$\Rightarrow 0(x-2)+\frac{7c}{4}(y-3)+c(z+4)=0$
$\begin{array}{rl}& \Rightarrow 0+\frac{7cy}{4}-\frac{21c}{4}+cz+4c=0\\ & \end{array}$
$\Rightarrow 7cy-21c+4cz+16c=0$
Dividing by c we have
$\begin{array}{rl}& 7y-21+4z+16=0\\ & \end{array}$
$\Rightarrow 7y+4z-5=0$

The Plane Excercise 28.11 Question 8

Answer:$x-19y-11z=0$
Hint: Use formula $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
Given: Points $(0,0,0)\mathrm{\&}(3,-1,2)$ and line $\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}$
Solution: We know that the equation of plane passing through $({\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1)$ is given by
$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ ……………….. (1)
So, equation of plane passing through (0,0,0) is
$a(x-0)+b(y-0)+c(z-0)=0$
$\begin{array}{rl}& \\ & ax+by+cz=0\end{array}$ …………………. (2)
It also passes through $(3,-1,2)$
So, equation (2) must satisfy the point $(3,-1,2)$
$3a-b+2c=0$ ………………… (3)
We know that line
$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ is parallel to plane $a_{2}x+b_{2}y+c_{2}z+d_2=0$ if
$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$ …………........ (4)
Here, the plane is parallel to line
$\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}$
So, $a\times 1+b\times (-4)+c\times 7=0$
$\begin{array}{rlr}& & \Rightarrow a-4b+7c=0\end{array}$ …………….… (5)
Solving equation (3) and (5) by cross multiplication we have,
$\begin{array}{rl}& \frac{\mathrm{a}}{-1\times 7-(-4)\times 2}=\frac{\mathrm{b}}{1\times 2-3\times 7}=\frac{\mathrm{c}}{3\times (-4)-1\times (-1)}\\ & \end{array}$
$\Rightarrow \frac{\mathrm{a}}{-7+8}=\frac{\mathrm{b}}{2-21}=\frac{\mathrm{c}}{-12+1}$
$\begin{array}{rl}& \Rightarrow \frac{a}{1}=\frac{b}{-19}=\frac{c}{-11}=k\\ & \end{array}$
$a=k,b=-19k,c=-11k$
Putting the value in equation (2) we get
$\begin{array}{rl}& ax+by+cz=0\\ \\ & \end{array}$
$kx-19ky-11kz=0$
Dividing by k we have
$x-19y-11z=0$
The required equation is $x-19y-11z=0$

The Plane Excercise 28.11 Question 9

Answer:$\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+k(-3\hat{i}+5\hat{j}+4\hat{k})$
or
$\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$
Hint: Use formula $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$
Given: Point $(1,2,3)$ and planes
$\begin{array}{rl}& \overrightarrow{r}\cdot (\hat{i}-\hat{j}+2\hat{k})=5\\ & \end{array}$
$\overrightarrow{r}\cdot (3\hat{i}+\hat{j}+2\hat{k})=6$
Solution: We know that, the equation of line passing through $(1,2,3)$ is given by
$\frac{x-1}{a_1}=\frac{y-2}{b_1}=\frac{z-3}{c_1}$ ……………….. (1)
We know that line
$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ is parallel to plane $a_{2}x+b_{2}y+c_{2}z+d_2=0$
If $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$ ………………… (2)
Here, line (1) is parallel to plane
$x-y+2z=5$
So, $a\times 1+b\times (-1)+c\times 2=0$
$\begin{array}{c}\Rightarrow a-b+2c=0\end{array}$ …………………. (3)
Also, line (1) is parallel to plane,
$3x+y+z=6$
So, $a\times 3+b\times (1)+c\times 1=0$
$\begin{array}{rlr}& & 3a+b+c=0\end{array}$ ………………….. (4)
Solving equation (3) and (4) by cross multiplication
We have,
$\begin{array}{rl}& \frac{a}{-1\times 1-1\times 2}=\frac{b}{3\times 2-1\times 1}=\frac{c}{1\times 1-3\times (-1)}\\ & \end{array}$
$\Rightarrow \frac{a}{-1-2}=\frac{b}{6-1}=\frac{c}{1+3}$
$\Rightarrow \frac{a}{-3}=\frac{b}{5}=\frac{c}{4}=k$
$a=-3k,b=5k,c=4k$
Putting the value in equation (1) we have
$\frac{x-1}{-3k}=\frac{y-2}{5k}=\frac{z-3}{4k}$
Multiplying by k we have
$\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$
The required equation is
$\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$
or
$\begin{array}{rlrl}& & & \overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+k(-3\hat{i}+5\hat{j}+4\hat{k})\end{array}$

The Plane Exercise 28.11 Question 10

Answer: The line of section is parallel to plane
Hint: Use properties of plane
Given:
$\begin{array}{rl}& 5x+2y-4z+2=0,2x+8y+2z-1=0\\ \\ & \mathrm{\&}4x-2y-5z-2=0\end{array}$
Solution: Let $a_1,b_1\mathrm{\&}{c}_1$ be the direction ratios of line $5x+2y-4z+2=0,2x+8y+2z-1=0$
As we know, that if two planes are perpendicular with direction ratios as $a_1,b_1\mathrm{\&}{c}_1$ and $a_2,b_2\mathrm{\&}{c}_2$ then $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
Since line lies in both the planes, so it is perpendicular to both planes
$\begin{array}{rl}& 5a_1+2b_1-4c_1=0\\ & \end{array}$ …………….. (1)
$2a_1+8b_1+2c_1=0$ …………….. (2)
Solving equation (1) and (2) by cross multiplication
We have,
$\begin{array}{rl}& \frac{a_1}{2\times 2-(-4)\times 8}=\frac{b_1}{2\times (-4)-5\times 2}=\frac{c_1}{5\times 8-2\times 2}\\ & \end{array}$
$\Rightarrow \frac{a_1}{4+32}=\frac{b_1}{-8-10}=\frac{c_1}{40-4}$
$\begin{array}{rl}& \Rightarrow \frac{a_1}{36}=\frac{b_1}{-18}=\frac{c_1}{36}=k\\ & \end{array}$
$\Rightarrow \frac{a_1}{2}=\frac{b_1}{-1}=\frac{c_1}{2}=k$
$a=2k,b=-k,c=2k$
We know that line
$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ is parallel to plane $a_{2}x+b_{2}y+c_{2}z+d_2=0$
if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$ ………….. (3)
Here, line with direction ratios is parallel to plane
$\begin{array}{rl}& 4x-2y-5z-2=0\\ & \end{array}$
$\Rightarrow 2\times 4+(-1)\times (-2)+2\times (-5)=0$
$\Rightarrow 8+2-10=0$
Therefore, the line of section is parallel to the plane.

The Plane Exercise 28.11 Question 11

Answer:
$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}})+\mathrm{k}((2\hat{\mathrm{i}}-\hat{\mathrm{j}}+3\hat{\mathrm{k}}))$
Hint: Use properties of plane
Given: Point $(1,-1,2)$ and plane $2x-y+3z-5=0$
Solution: Equation of line passing through $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\overrightarrow{r}=\overrightarrow{a}+k\overrightarrow{b}$ ……………. (1)
Given that the line passes through $(1,-1,2)\text{ is }\overrightarrow{r}=(\hat{i}-\hat{j}+2\hat{k})+k\overrightarrow{b}$ …………….. (2)
Since, line (1) is perpendicular to the plane $2x-y+3z-5=0$
So, normal to plane is parallel to the line.
In vector form,
$\overrightarrow{\mathrm{b}}=\mathrm{m}(2\hat{\mathrm{i}}-\hat{\mathrm{j}}+3\hat{\mathrm{k}})$ as, is any scalar.
Thus, the equation of required line,
$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}})+\mathrm{k}((2\hat{\mathbf{i}}-\hat{\mathrm{j}}+3\hat{\mathrm{k}}))$

The Plane Exercise 28.11 Question 12

Answer: $12x+15y-14z-68=0$
Hint: Use formula $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
Given: Points $(2,2,-1)\mathrm{\&}(3,4,2)$ ratios $7,0,6$
Solution: We know that the equation of plane passing through $({\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1)$ is
$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ ……………….. (1)
So, equation of plane passing through $(2,2,-1)$ is
$a(x-2)+b(y-2)+c(z+1)=0$ ………………… (2)
It also passes through $(3,4,2)$
So, equation (2) passes through $(3,4,2)$
$\begin{array}{rl}& a(3-2)+b(4-2)+c(2+1)=0\\ & \end{array}$
$\Rightarrow a+2b+3c=0$ …………………. (3)
We know that line
$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ is parallel to plane $a_{2}x+b_{2}y+c_{2}z+d_2=0$
if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$ …………… (4)
Here, the plane (2) is parallel to line having direction ratios 7, 0 ,6
So, $\begin{array}{rl}& a\times 7+b\times 0+c\times 6=0\\ & \end{array}$
$\Rightarrow 7a+6c=0$
$\Rightarrow a=\frac{-6c}{7}$ ………………… (5)
Putting the value of a in equation (3) we get
$\begin{array}{rl}& a+2b+3c=0\\ & \end{array}$
$\Rightarrow \frac{-6c}{7}+2b+3c=0$
$\Rightarrow 14b+15c=0$
$\Rightarrow b=\frac{-15c}{14}$
Putting the value of a and b in equation (2) we get
$\begin{array}{rl}& a(x-2)+b(y-2)+c(z+1)=0\\ & \end{array}$
$\Rightarrow \frac{-6c}{7}(x-2)+\frac{-15c}{14}(y-2)+c(z+1)=0$
$\Rightarrow \frac{-6xc}{7}+\frac{12c}{7}-\frac{15cy}{14}+\frac{30c}{14}+cz+c=0$
Multiplying by $\frac{14}{c}$ we have
$\begin{array}{rl}& \Rightarrow -12x+24-15y+30+14z+14=0\end{array}$
$\Rightarrow -12x-15y+14z+68=0$
$\Rightarrow 12x+15y-14z-68=0$
Equation of required plane is $12x+15y-14z-68=0$

The Plane exercise 28.11 question 13

Answer:${\sin }^{-1}\left(\frac{\sqrt{7}}{\sqrt{52}}\right)$
Hint: Use formula $\sin \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
Given: Line is $\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2}$ and plane is $3x+4y+z+5=0$
Solution: We know that line
$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ is parallel to plane $a_{2}x+b_{2}y+c_{2}z+d_2=0$
If $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
and the angle between them is given by
$\sin \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$ …………….. (1)
Now, given equation by line is
$\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2}$
So, $a_1=3,b_1=-1,c_1=2$
Equation of plane is $3x+4y+z+5=0$
So, $\begin{array}{rl}{a}_2=& 3,b_2=4,c_2=1\mathrm{\&}{d}_2=-5\end{array}$
$\sin \theta =\frac{3\times 3+(-1)\times 4+2\times 1}{\sqrt{3^2+(-1{)}^2+2^2}\sqrt{3^2+4^2+1^2}}$
$\sin \theta =\frac{9-4+2}{\sqrt{9+1+4}\sqrt{9+16+1}}$
$\begin{array}{rl}& =\frac{7}{\sqrt{14}\sqrt{26}}\times \frac{\sqrt{7}}{\sqrt{7}}\\ & \end{array}$
$=\frac{7\sqrt{7}}{7\sqrt{52}}=\frac{\sqrt{7}}{\sqrt{52}}$
$\sin \theta =\frac{\sqrt{7}}{\sqrt{52}}\Rightarrow \theta ={\sin }^{-1}\left(\frac{\sqrt{7}}{\sqrt{52}}\right)$
The angle between the plane and the line is $\theta ={\sin }^{-1}\left(\frac{\sqrt{7}}{\sqrt{52}}\right)$

The Plane exercise 28.11 question 14

Answer: Therefore the equation of the plane is $9x-8y+7z-21=0$ and distance is $\mathrm{D}=\frac{13}{\sqrt{194}}$
Hint: Use formula $ax+by+cz+d=0$ and $D=\frac{a{x}_1+b{y}_1+c{z}_1+d_1}{\sqrt{a^2+b^2+c^2}}$
Given: Planes are $x-2y+z=1\mathrm{\&}2x+y+z=8$ and point $(1,1,1)$ ratios proportional is $1,2,1.$
Solution: We know that equation of plane passing through the intersection of planes $a_{1}x+b_{1}y+c_{1}z+d_1=0\text{ and }{a}_{2}x+b_{2}y+c_{2}z+d_2=0$ is given by
$(a_{1}x+b_{1}y+c_{1}z+d_1)+k(a_{2}x+b_{2}y+c_{2}z+d_2)=0$
So, equation of plane passing through the intersection of planes $x-2y+z-1=0\mathrm{\&}2x+y+z-8=0$ is
$\begin{array}{rl}& \Rightarrow (x-2y+z-1)+k(2x+y+z-8)=0\\ & \end{array}$ ………………. (1)
$\Rightarrow x(1+2k)+y(-2+k)+z(1+k)+(-1-8k)=0$
We know that line
$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$
Is parallel to plane $a_{2}x+b_{2}y+c_{2}z+d_2=0\text{ if }{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0$
Given the plane is parallel to line with direction ratios 1, 2, 1
$\begin{array}{rl}& 1\times (1+2k)+2\times (-2+k)+1\times (1+k)=0\\ & \end{array}$
$\Rightarrow 1+2k-4+2k+1+k=0$
$\Rightarrow k=\frac{2}{5}$
Putting the value of k in equation (1) we get
$\begin{array}{rl}& \Rightarrow (x-2y+z-1)+\frac{2}{5}(2x+y+z-8)=0\\ & \end{array}$
$\Rightarrow x+\frac{4}{5}x-2y+\frac{2}{5}y+z+\frac{2}{5}z-1-\frac{16}{5}=0$
$\Rightarrow \frac{9x}{5}-\frac{8y}{5}+\frac{7z}{5}-\frac{21}{5}=0$
$\Rightarrow 9x-8y+7z-21=0$
We know that the distance (D) of point $({\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1)$ from plane $ax+by+cx-d=0$ is given by
$D=\frac{a_{1}x+b_{1}y+c_{1}z+d_1}{\sqrt{a^2+b^2+c^2}}$
So, distance of Point $(1,1,1)$ from plane is
$\begin{array}{rl}D& =\frac{9\times 1+(-8)\times 1+7\times 1-21}{\sqrt{9^2+(-8{)}^2+7^2}}\end{array}$
$=\frac{9-8+7-21}{\sqrt{81+64+49}}$
$=\frac{-13}{\sqrt{194}}$
Taking the mod value we have $D=\frac{13}{\sqrt{194}}\text{ unit }$