JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Three Dimensional Geometry

The given line passes through $(1,3,4)$ with direction $\vec{d}=(2,1,2)$, and the plane has normal $\vec{n}=(1,-2,-1)$.
Since $\vec{d}\cdot \vec{n}=-2\ne 0$, the line meets the plane, and that common point lies on the projection $L$. Substituting $(1+2t,3+t,4+2t)$ in the plane equation gives $-9-2t=3$, so $t=-6$ and the point is $A(-11,-3,-8)$.
The direction of $L$ is the part of $\vec{d}$ lying in the plane:

so $\vec{u}=(7,1,5)$ may be used.
With $Q(0,0,6)$ one has $\overrightarrow{AQ}=(11,3,14)$ and $\overrightarrow{AQ}\times \vec{u}=(1,43,-10)$, so