JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Three Dimensional Geometry

The first line $x=ay-1=z-2$ passes through $\left(0,\frac{1}{a},2\right)$ with direction $\left(1,\frac{1}{a},1\right)$, and the second line $x=3y-2=bz-2$ passes through $\left(0,\frac{2}{3},\frac{2}{b}\right)$ with direction $\left(1,\frac{1}{3},\frac{1}{b}\right)$. Coplanarity requires the scalar triple product of the joining vector and the two directions to vanish:

Expanding along the first row and using $\frac{2}{b}-2=2\left(\frac{1}{b}-1\right)$,

Since $a\ne 0$, the condition is $\frac{1}{b}=1$. The lines are coplanar when $b=1$, with $a$ any non-zero real number.