Question

The line of shortest distance between the lines $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}\text{and}\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$ 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 $P\text{is}(\alpha ,\beta ,\gamma )$, then $\alpha +\beta -\gamma$ is equal to

Answer 1

Note: It is a bonus question
Reason: On solving, $a$ is coming out to be imaginary.
Line of shortest distance will be along $\overline{b_1}\times \overline{b_2}$

Where, $\overline{b_1}=\hat{j}+\hat{k}\text{and}{\overrightarrow{b}}_2=2\hat{i}+2\hat{j}+\hat{k}$

$\overline{b_1}\times \overline{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& 1\\ 2& 2& 1\end{array}\right|=-\hat{i}+2\hat{j}-2\hat{k}$

Angle between $\overline{b_1}\times \overline{b_2}\text{and plane}P$,

$\sin \theta =\left|\frac{-a-2+2}{3\cdot \sqrt{a^2+2}}\right|=\frac{5}{\sqrt{27}}\Rightarrow \frac{|a|}{\sqrt{a^2+2}}=$

$\Rightarrow a^2=-\frac{25}{11}\text{(not possible)}$