Given,

$P$ be the plane passing through the line $\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$ and the point $(2,4,-3)$,

So, equation of plane is given by,

$\left|\begin{array}{ccc}x-2& y-4& z+3\\ 1& -3& 7\\ 2-1& 4-2& -3-\left(-5\right)\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x-2& y-4& z+3\\ 1& -3& 7\\ 1& 2& 2\end{array}\right|=0$

$\Rightarrow \left(x-2\right)\left(-20\right)-\left(y-4\right)\left(-5\right)+\left(z+3\right)\left(5\right)=0$

$\Rightarrow 4 x-y-z-7=0$

Now finding the image$\left(\alpha , \beta , \gamma \right)$  of the point $\left(-1,3,4\right)$ with respect to plane $4 x-y-z-7=0$ we get,

$\frac{x+1}{4}=\frac{y-3}{-1}=\frac{z-4}{-1}=-2\left(\frac{-4-3-4-7}{16+1+1}\right)=2$

$\Rightarrow x=7=\alpha , y=1=\beta , z=2=\gamma$

Hence, $\alpha +\beta +\gamma =10$