If for some

Question:

If for some $\alpha \in \mathbf{R}$, the lines $L_{1}: \frac{x+1}{2}=\frac{y-2}{-1}=\frac{z-1}{1}$ and

$L_{2}: \frac{x+2}{\alpha}=\frac{y+1}{5-\alpha}=\frac{z+1}{1}$ are coplanar, then the line $L_{2}$ passes through the point :

  1. $(10,2,2)$

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

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

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


Correct Option: , 2

Solution:

Since, lince are coplanar

$\therefore\left|\begin{array}{ccc}1 & 3 & 2 \\ 2 & -1 & 1 \\ \alpha & 5-\alpha & 1\end{array}\right|=0$

$\Rightarrow 1(-1-5+\alpha)-3(2-\alpha)+2(10-2 \alpha+\alpha)=0$

$\therefore \alpha=-4$

$\therefore$ Equation of $L_{2}: \frac{x+2}{-4}=\frac{y+1}{9}=\frac{z+1}{1}$

$\therefore$ Point $(2,-10,-2)$ lies on line $L_{2}$.

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