JEE Main 2020 (Online) 4th September Evening ShiftQuestion 1
The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line $\frac x2=\frac y3=\frac z{-6}$ is :
3D Geometry
Solution
Point along the line: $(1+2u,-2+3u,3-6u)$. On plane: $(1+2u)-(-2+3u)+(3-6u)=6-7u=5\Rightarrow u=\tfrac17$. Distance $=|u|\sqrt{4+9+36}=\tfrac17\cdot7=1$. Answer (B).