Question

Let $P$ be a plane containing the line $\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+5}{2}$ and parallel to the line $\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z+5}{7}$. If the point $(1,–1,\alpha )$ lies on the plane $P$, then the value of $\left|5\alpha \right|$ is equal to

Answer 1

Let $L_1:\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+5}{2}$ and $L_2:\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z+5}{7}$
The plane contains the points
$(1,-6,-5)$ and $(1,-1,\alpha )$
Vector joining both points is
$\overrightarrow{V}=5\hat{j}+(\alpha +5)\hat{k}$
So, $\overrightarrow{V}\cdot \overrightarrow{n}=0$, where $\overrightarrow{n}$ is normal vector to the plane
Now,
$\left|\begin{array}{ccc}0& 5& \alpha +5\\ 3& 4& 2\\ 4& -3& 7\end{array}\right|=0\Rightarrow -5\left(13\right)+(\alpha +5)(-25)=0\Rightarrow -13-25-5a=0\Rightarrow 5a=-38\therefore |5a|=38$