Given a line with direction ratios $a,-4 a,-7$ be perpendicular to the lines 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 $\left(\alpha ,\beta ,\gamma \right)$, then $\alpha +\beta +\gamma$ is equal to ________.

Now given $\left(a,-4a,-7\right)\perp$ to $\left(3,-1,2b\right)$

So, $3a+4a-14b=0$

$\Rightarrow a=2b \cdots \left(\mathrm{i}\right)$

Also $\left(a,-4a,-7\right)\perp$ to $\left(b,a,-2\right)$

So, $ab-4a^2+14=0 \cdots \left(\mathrm{ii}\right)$

From Equations $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$ we get,

$2b^2-16b^2+14=0$

$\Rightarrow b^2=1$

So, $a^2=4b^2=4$

So, line becomes $\frac{x+1}{5}=\frac{y-2}{3}=\frac{z}{1}=k$

Let points be $\alpha =5k-1,\beta =3k+2,\gamma =k$

As $\left(\alpha ,\beta ,\gamma \right)$ satisfies plane $x-y+z=0$

So, $5k-1-\left(3k+2\right)+k=0$

$\Rightarrow k=1$

$\therefore \alpha +\beta +\gamma =9k+1=10$