JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Three Dimensional Geometry

A line through $(0,1,2)$ with direction ratios $(p,q,r)$ is perpendicular to the given line, whose direction ratios are $(2,3,-2)$, exactly when $2p+3q-2r=0$. Among the directions offered, $(-3,4,3)$ satisfies this, since $-6+12-6=0$, while $(3,4,3)$, $(3,4,-3)$ and $(3,-4,3)$ give $12$, $24$ and $-12$. The required line is $\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$.