Question

If the equation of plane passing through the mirror image of a point $(2,3,1)$ with respect to line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$ and containing the line $\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}$ is $\alpha x+\beta y+\gamma z=24,$ then $\alpha +\beta +\gamma$ is equal to :

• A. $21$
• B. $19$
• C. $18$
• D. $20$

Answer 1

The correct option is B $19$


Let $L_1=\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$
Let point $M\equiv (2\lambda -1,\lambda +3,-\lambda -2)$
Direction ratios of AM line are $(2\lambda -1-2,\lambda +3-3,-\lambda -2-1)$
$\equiv (2\lambda -3,\lambda ,-\lambda -3)$
$\because AM$ is $\perp$ line to $L_1$
$\Rightarrow 2(2\lambda -3)+1\left(\lambda \right)-1(-\lambda -3)=0$
$\Rightarrow \lambda =\frac{1}{2}$
$\therefore M\equiv (0,\frac{7}{2},\frac{-5}{2})$
$\because M$ is mid point of $A$ and $B$
$\Rightarrow M=\frac{A+B}{2}$
$\Rightarrow B=2M-A$
$\Rightarrow B\equiv (-2,4,-6)$
Now we have to find equation of plane passing through $B(-2,4,-6)$ and also containing the line
$\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}\cdots \left(1\right)$
$\frac{x-2}{3}=\frac{y-1}{-2}=\frac{z+1}{1}$
Point $P$ on the line is $(2,1,-1)$
$d.r.s$ of $\overrightarrow{b_2}$ of the line $L_2$ is $(3,-2,1)$
$\overrightarrow{n}$ is $\parallel$ to $(\overrightarrow{b_2}\times \overrightarrow{PB})$
$\overrightarrow{b_2}=3\hat{i}-2\hat{j}+\hat{k}$
$\overrightarrow{PB}=-4\hat{i}+3\hat{j}-5\hat{k}$
$\overrightarrow{n}=7\hat{i}+11\hat{j}+\hat{k}$
$\therefore$ equation of plane is $\overrightarrow{r}\cdot \overrightarrow{n}=\overrightarrow{a}\cdot \overrightarrow{n}$
$\overrightarrow{r}\cdot (7\hat{i}+11\hat{j}+\hat{k})=(-2\hat{i}+4\hat{j}-6\hat{k})\cdot (7\hat{i}+11\hat{j}+\hat{k})$
$7x+11y+z=-14+44-6$
$7x+11y+z=24$
$\therefore \alpha +\beta +\gamma =19$