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 $P:ax-y-z=0$, $(a>0)$. 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

Answer: 3

DR's of line of shortest distance
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& 1\\ 2& 2& 1\end{array}\right|=-\hat{i}+2\hat{j}-2\hat{k}$
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 $(1,-1,-1)$
$\sin \alpha =\left|\frac{-a-2+2}{\sqrt{4+4+1}\sqrt{a^2+1+1}}\right|=\frac{5}{3\sqrt{3}}$
$\sqrt{3}|a|=5\sqrt{a^2+2}$
$3a^2=25a^2+50$
No value of (a)