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

DR’s of line of shortest distance

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

angle between line and plane is ${\cos }^{-1}\sqrt{\frac{2}{27}}=\mathrm{\alpha }$

$\cos \mathrm{\alpha }=\sqrt{\frac{2}{27}},\sin \mathrm{\alpha }=\frac{5}{3\sqrt{3}}$

DR’s normal to plane (a, –1, –1)

$\begin{array}{l}\cos \alpha =\left|\frac{-a-2+2}{\sqrt{4+4+1}\sqrt{a^2+1+1}}\right|=\frac{\sqrt{2}}{3\sqrt{3}}\\ \frac{2}{27}=\frac{a^2}{9\left(a^2+2\right)}\Rightarrow a=\pm 2\\ for a=2\\ P:2x-y-z=0\\ The image of P\left(1,1,-5\right) to the plane is\left(\alpha ,\beta ,\gamma \right)\\ \frac{\alpha -1}{2}=\frac{\beta -1}{-1}=\frac{\gamma -1}{-1}=\frac{-2\left(2-1+5\right)}{6}\\ \Rightarrow \alpha =-3,\beta =3,\gamma =-3\\ \therefore \alpha +\beta -\gamma =3\end{array}$

No value of (a)