The line of shortest distance between the lines x - 2 0 = y - 1 1 = z 1 and x - 3 2 = y - 5 2 = z - 1 1…

Given lines, $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$ and $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$

Now direction ratio of line of shortest distance will be, 

$\left|\begin{array}{lll}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& 1\\ 2& 2& 1\end{array}\right|=-\hat{i}+2\hat{j}-2\hat{k}$

Now given angle between line and plane is ${\sin }^{-1}\sqrt{\frac{2}{27}}=\alpha$, so $\sin \alpha =\sqrt{\frac{2}{27}}$

Now direction ratio of normal to plane is $\left(a,-1,-1\right)$

So, $\sin \alpha =\left|\frac{-a-2+2}{\sqrt{4+4+1}\sqrt{a^2+1+1}}\right|=\frac{\sqrt{2}}{3\sqrt{3}}$

$\Rightarrow \sqrt{3}\left|a\right|=\sqrt{2}\sqrt{a^2+2}$

$\Rightarrow 3{\mathrm{a}}^2=2{\mathrm{a}}^2+4$

$\Rightarrow {\mathrm{a}}^2=4$

$\Rightarrow a=2 \left(\text{as} a>0 \text{given}\right)$

Now equation of plane will be $2x-y-z=0$ and image of point $\left(1,1,-5\right)$ will be,

$\frac{x-1}{2}=\frac{y-1}{-1}=\frac{z+5}{-1}=-2\frac{\left(2-1+5\right)}{2^2+1^2+1^2}$

$\Rightarrow \frac{x-1}{2}=\frac{y-1}{-1}=\frac{z+5}{-1}=-2$

$\Rightarrow x=-3, y=3, z=-3$

So, $\alpha =-3, \beta =3, \gamma =-3$

So, $\alpha +\beta -\gamma =-3+3+3=3$

 

Note: This question was bonus in JEE mains 25th july 2022 paper, we have done some modification to the original question.