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 $\mathrm{P}$, then the value of $|5 \alpha|$ is equal to___________
Solution:
Equation of plane is $\left|\begin{array}{ccc}\mathrm{x}-1 & \mathrm{y}+6 & \mathrm{z}+5 \\ 3 & 4 & 2 \\ 4 & -3 & 7\end{array}\right|=0$
Now $(1,-1, \alpha)$ lies on it so
$\left|\begin{array}{ccc}0 & 5 & \alpha+5 \\ 3 & 4 & 2 \\ 4 & -3 & 7\end{array}\right|=0 \Rightarrow 5 \alpha+38=0 \Rightarrow|5 \alpha|=38$