JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Three Dimensional Geometry

The required plane holds the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$, direction ${\vec{d}}_1=(2,3,4)$, and must be perpendicular to the plane of the lines with directions $(3,4,2)$ and $(4,2,3)$. The normal of that second plane is $(3,4,2)\times (4,2,3)=(8,-1,-10)$. The wanted plane contains ${\vec{d}}_1$ and has this vector lying in it, so its own normal is ${\vec{d}}_1\times (8,-1,-10)=(26,-52,26)$, i.e. $(1,-2,1)$ after dividing by $26$. Since the line passes through the origin, the plane is $x-2y+z=0$.