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Q. Let the image of the point $P(2,-1,3)$ in the plane $x+2 y-z=0$ be $Q$. Then the distance of the plane $3 x+2 y+z+29=0$ from the point $Q$ is

JEE MainJEE Main 2023Three Dimensional Geometry

Solution:

image
eq. of line PM $ \frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-1}=\lambda$
any point on line$=(\lambda+2,2 \lambda-1,-\lambda+3)$
for point ' $ m \text { ' }(\lambda+2)+2(2 \lambda-1)-(3-\lambda)=0$
$ \lambda=\frac{1}{2}$
$ \text { Point } m \left(\frac{1}{2}+2,2 \times \frac{1}{2}-1, \frac{-1}{2}+3\right)$
$ =\left(\frac{5}{2}, 0, \frac{5}{2}\right)$
For Image $Q (\alpha, \beta, \gamma)$
$\frac{\alpha+2}{2}=\frac{5}{2}, \frac{\beta-1}{2}=0 $
$ \frac{\gamma+3}{2}=\frac{5}{2} $
$ Q:(3,1,2) $
$ d =\left|\frac{3(3)+2(1)+2+29}{\sqrt{3^2+2^2+1^2}}\right| $
$ d =\frac{42}{\sqrt{14}}=3 \sqrt{14}$