JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Three Dimensional Geometry

The direction ratios of the line of intersection are given by the cross product of normals: $(bc,-ac,0)$, which simplifies to direction ratios $(b,-a,0)$. The angle between the line and the plane $y-z+2=0$ (with normal $(0,1,-1)$) is $30^{\circ }$, so the angle with the normal is $60^{\circ }$. Using $\cos 60^{\circ }=\frac{|-a|}{\sqrt{2}\sqrt{a^2+b^2}}=\frac{1}{2}$, we get $b=\pm a$. Taking $b=a$, the direction ratios are $(a,-a,0)$, giving direction cosines $\left(\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0\right)$.