Given,

The equation of the plane passing through the line of intersection of the planes $2x-y+z=3, 4x-3y+5z+9=0$ and parallel to the line $\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$ is $ax+by+cz+6=0,$

Now let, $P_1=2x-y+z=3$

And $P_2=4x-3y+5z+9=0$

Now using family of planes we get,

$P_1+\lambda P_2=0$

$\Rightarrow (2x-y+z-3)+\lambda (4x-3y+5z+9)=0$

$\Rightarrow (2+4\lambda )x-(1+3\lambda )y+(1+5\lambda )z-(3-9\lambda )=0$

Now let,

$P_3=(2+4\lambda )x-(1+3\lambda )y+(1+5\lambda )z-(3-9\lambda )=0$

Now given $P_3$ is parallel to line,

$\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$

So, using perpendicular condition as normal vector of plane will be perpendicular to given line we get,

$-2(2+4\lambda )-4(1+3\lambda )+5(1+5\lambda )=0$

$\Rightarrow -3+5\lambda =0$

$\Rightarrow \lambda =\frac{3}{5}$

Hence, equation of plane $P_3$ will be,

$P_3:\frac{22x}{5}-\frac{14y}{5}+\frac{20z}{5}+\frac{12}{5}=0$

$\Rightarrow P_3=11x-7y+10z+6=0$

Now on comparing with $ax+by+cz+6=0,$ we get,

$a=11$, $b=-7$ and $c=10$

$\therefore a+b+c=11-7+10=14$