JEE Main 2023 (Online) 1st February Morning ShiftQuestion 1
Let the image of the point $P(2,-1,3)$ in the plane $x+2y-z=0$ be $Q$. Then the distance of the plane $3x+2y+z+29=0$ from the point $Q$ is
3D Geometry
Solution
$Q=P-2\frac{(2-2-3)}{6}(1,2,-1)=(3,1,2)$. Distance from $3x+2y+z+29=0$: $\frac{|9+2+2+29|}{\sqrt{14}}=3\sqrt{14}$.