JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Three Dimensional Geometry

The line through $(1,-2,3)$ and $(1,1,0)$ has direction vector $(0,1,-1)$, so a general point on it is $(1,-2+t,3-t)$. The vector joining the given point $(4,2,3)$ to this point is $(-3,t-4,-t)$. For the foot of the perpendicular this vector must be orthogonal to the line's direction, so $(0)(-3)+(1)(t-4)+(-1)(-t)=0$, giving $2t-4=0$ and $t=2$. The foot is therefore $(1,0,1)$. Checking the candidate planes, $2x+y-z=2+0-1=1$, so the foot lies on the plane $2x+y-z=1$.