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  $\frac{\mathrm{x}+1}{-2}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}-2}{5}$is $\mathrm{a} \mathrm{x}+\mathrm{b} \mathrm{y}+\mathrm{c} \mathrm{z}+6=0$, then $a + b + c$is equal to

${\mathrm{\pi }}_1=2\mathrm{x}-\mathrm{y}+\mathrm{z}=0,$                        ${\mathrm{\pi }}_2=4\mathrm{x}-\mathrm{y}+5\mathrm{z}+9=0$

Any plane passing through line of intersection is given by ${\mathrm{\pi }}_1+\lambda {\pi }_2=0$

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

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

Given the plane is parallel to line $\frac{\mathrm{x}+1}{-2}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}-2}{5}$

$\therefore -4-8\mathrm{\lambda }-4-12\mathrm{\lambda }+5+25\mathrm{\lambda }=0\Rightarrow \mathrm{\lambda }=\frac{3}{5}$

$\left(\frac{22}{5}\right)x-\left(\frac{14}{5}\right)y+\left(\frac{20}{5}\right)z+\frac{12}{5}=0\Rightarrow 11k-7y+10z+6=0$

$a = 11, b = -7, c = 10 a + b + c = 14$