Question

The line of shortest distance between the lines $ \frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1} $ and $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$ makes an angle of $\cos ^{-1}$ $\left(\sqrt{\frac{2}{27}}\right)$ with the plane $P: a x-y-z=0$, $(a>0)$. If the image of the point $(1,1,-5)$ in the plane $P$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta-\gamma$ is equal to ______

Answer 1

DR's of line of shortest distance
$\begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\0 & 1 & 1 \\2 & 2 & 1\end{vmatrix}=-\hat{ i }+2 \hat{ j }-2 \hat{ k }$
angle between line and plane is $\cos ^{-1} \sqrt{\frac{2}{27}}=\alpha$
$\cos \alpha=\sqrt{\frac{2}{27}}, \sin \alpha=\frac{5}{3 \sqrt{3}}$
DR's normal to plane $(1,-1,-1)$
$\sin \alpha=\left|\frac{-a-2+2}{\sqrt{4+4+1} \sqrt{a^2+1+1}}\right|=\frac{5}{3 \sqrt{3}} $
$ \sqrt{3}| a |=5 \sqrt{a^2+2}$
$ 3 a^2=25 a^2+50$
No value of (a)