If for some $\alpha \in \mathrm{R}$, the lines
$\mathrm{L}_{1}: \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-1}{1}$ and
$\mathrm{L}_{2}: \frac{\mathrm{x}+2}{\alpha}=\frac{\mathrm{y}+1}{5-\alpha}=\frac{\mathrm{z}+1}{1}$ are coplanar, then the
line $L_{2}$ passes through the point :
Correct Option: , 4
$\mathrm{L}_{1} \equiv \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-1}{1}$
$\mathrm{L}_{2} \equiv \frac{\mathrm{x}+2}{\alpha}=\frac{\mathrm{y}+1}{5-\alpha}=\frac{\mathrm{z}+1}{1}$
Point $\mathrm{A}(-1,2,1) \mathrm{B}(-2,-1,-1)$
$\because \mathrm{L}_{1}$ and $\mathrm{L}_{2}$ are coplanar
$\Rightarrow\left|\begin{array}{ccc}2 & -1 & 1 \\ \alpha & 5-\alpha & 1 \\ 1 & 3 & 2\end{array}\right|=0$
$\alpha=-4$
$\mathrm{L}_{2} \equiv \frac{\mathrm{x}+2}{-4}=\frac{\mathrm{y}+1}{9}=\frac{\mathrm{z}+1}{1}$
Check options $(2,-10,-2)$ lies on $\mathrm{L}_{2}$