JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Three Dimensional Geometry

The given line passes through $(-3,1,5)$ with direction $(-3,1,5)$. A second line is coplanar with it when $(\overrightarrow{P_2}-\overrightarrow{P_1})\cdot (\overrightarrow{d_1}\times \overrightarrow{d_2})=0$. Take the line through $(-1,2,5)$ with direction $(-1,2,5)$, so $\overrightarrow{P_2}-\overrightarrow{P_1}=(2,1,0)$. Here $\overrightarrow{d_1}\times \overrightarrow{d_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|=(-5,10,-5)$. Then $(2,1,0)\cdot (-5,10,-5)=-10+10+0=0$, so this line is coplanar with the given line.