If the equation of the plane passing through the line of intersection of the planes $2 x-y+z=3,4 x-3 y+5 z+9=0$ and parallel to the line $\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$ is $a x+b y+c z+6=0$, then $a+b+c$ is equal to :

Equation of plane intersection of two plane <br/><br/>$\mathrm{P}_1+\lambda \mathrm{P}_2=0$ <br/><br/>$\mathrm{P}_1: 2 x-y+z=3, \text { and } \mathrm{P}_2: 4 x-3 y+5 z+9=0$ <br/><br/>Equation of any plane passing through the intersection of given planes is <br/><br/>$(2 x-y+z-3)+\lambda(4 x-3 y+5 z+9)=0$ <br/><br/>$\Rightarrow(2+4 \lambda) x+(-1-3 \lambda) y+(1+5 \lambda) z+(-3+9 \lambda)=0$ ..........(i) <br/><br/>Plane (i) is parallel to given line <br/><br/>$$ \begin{aligned} & \therefore (-2)(2+4 \lambda)+4(-1-3 \lambda)+5(1+5 \lambda) =0 \\\\ & \Rightarrow -4-8 \lambda-4-12 \lambda+5+25 \lambda =0 \\\\ & \Rightarrow 5 \lambda-3 =0 \\\\ & \Rightarrow \lambda =\frac{3}{5} \end{aligned} $$ <br/><br/>Putting the value of $\lambda$ in Eq. (i), we get <br/><br/>$$ \begin{aligned} & (2 x-y+z-3)+\frac{3}{5}(4 x-3 y+5 z+9)=0 \\\\ & \Rightarrow 11 x-7 y+10 z+6=0 \\\\ & \therefore a=11, b=-7, c=10 \text { and } d=6 \\\\ & \therefore a+b+c=11-7+10=14 \end{aligned} $$