# If for some α ∈ R, the lines

Question:

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 :

1. $(-2,10,2)$

2. $(10,2,2)$

3. $(10,-2,-2)$

4. $(2,-10,-2)$

Correct Option: , 4

Solution:

$\mathrm{L}_{1} \equiv \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-1}{1}$

$\mathrm{L}_{2} \equiv \frac{\mathrm{x}+2}{\alpha}=\frac{\mathrm{y}+1}{5-\alpha}=\frac{\mathrm{z}+1}{1}$

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 $\mathrm{L}_{2}$