Let $a$ line with direction ratios $a,-4a,-7$ be perpendicular to the line with direction ratios $3,-1,2b$ and $b,a,-2$. If the point of intersection of the line  $\frac{x+1}{a^2+b^2}=\frac{y-2}{a^2-b^2}=\frac{z}{1}$ and the plane  $x-y+z=0$ is  $(\alpha ,\beta ,\gamma )$, then $\alpha +\beta +\gamma$  is equal to ___ 

Since D.R.s of (a, -4a, -7) are perpendicular to 3, -1, 2b
$(a,-4a,-7)\perp to(3,-1,2)$
a = 2b
$(a,-4a,-7)\perp to(b,a,-2)$
$3a+4a-14b=0$
$ab-4a^2+14=0$
From equations(1) and  $2b^2-16b^2+14=0$
$b^2=1$
$a^2=4b^2=4$
$\frac{x+1}{5}=\frac{y-2}{3}=\frac{z}{1}=k$
$=5k-1,\beta =3k+2,\gamma =k$
As $(\alpha ,\beta ,\gamma )$  satisfies x – y + z = 0
$5k-1-(3k+2)+k=0\Rightarrow k=1$
$\therefore \alpha +\beta +\gamma =9k+1=10$