Given plane equations are 

$3x-2y+4z-7 = 0 \& x+5y-2z+9=0$

Family of planes $\Rightarrow P_1+\lambda P_2=0$

$\Rightarrow (3x-2y+4z-7)+\lambda (x+5y-2z+9)=0$
Passing through the point $(1,4,-3)$
$\Rightarrow (3-8-12-7)+\lambda (1+20+6+9)=0$

$\Rightarrow \lambda =\frac{24}{36}=\frac{2}{3}$

Required equation of plane is $\left(3+\frac{2}{3}\right)x+y\left(-2+\frac{10}{3}\right)+z\left(4-\frac{4}{3}\right)-7+6=0$

 $\Rightarrow -11x-4y-8z+3=0$

Given equation is $\alpha x+\beta y+\gamma z+3=0$

By comparing above equations

So, $\alpha +\beta +\gamma =-23$