JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Three Dimensional Geometry

The required plane holds the given line and its projection on $2x+3y-z=5$, which means it must be the plane through the line standing perpendicular to that plane. Its normal $(a,b,c)$ is therefore perpendicular both to the line's direction $(2,-1,3)$ and to the plane's normal $(2,3,-1)$: $2a-b+3c=0$ and $2a+3b-c=0$. Solving gives $b=c$ and $a=-c$, so the normal is $(-1,1,1)$, and through the line's point $(3,-2,1)$ the plane is $-(x-3)+(y+2)+(z-1)=0$, that is $x-y-z=4$. Testing the candidates, $(2,0,-2)$ gives $2-0+2=4$, so it lies on the plane.