Let image of the line $\frac{x-1}{3}=\frac{y-3}{5}=\frac{z-4}{2}$ in the plane $2x –y + z + 3 = 0$ be L. A plane $7x + By + Cz + 3D = 0$ is such that if contains the line L and perpendicular to the plane $2x–y+z+3=0$ then the value of $B + C + D =\_\_\_\_\_\_\_$

Equation of L is $\frac{x+3}{1}=\frac{y-5}{6}=\frac{z-4}{2}$

Equation of the plane which contains the line L is
$a(x + 3) + b(y – 5) + c(z – 2) = 0$
Also $a + 6b + c = 0 and 2a –b + c = 0$

By solving $\frac{a}{7}=\frac{b}{1}=\frac{c}{-13}$

Hence equation of required plane is
$7x + y –13z + 42 = 0$