Find the equation of the plane which is perpendicular to the plane $5x + 3y + 6z + 8 = 0$ and which contains the line of intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$.

(A) The equation of a plane passing through the line of intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ is given by $(x + 2y + 3z - 4) + \lambda(2x + y - z + 5) = 0$.
Rearranging the terms,we get: $x(1 + 2\lambda) + y(2 + \lambda) + z(3 - \lambda) + (5\lambda - 4) = 0 \quad \dots(i)$
Since this plane is perpendicular to the plane $5x + 3y + 6z + 8 = 0$,the dot product of their normal vectors must be zero $(a_1a_2 + b_1b_2 + c_1c_2 = 0)$:
$5(1 + 2\lambda) + 3(2 + \lambda) + 6(3 - \lambda) = 0$
$5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0$
$7\lambda + 29 = 0 \implies \lambda = -\frac{29}{7}$
Substituting $\lambda = -\frac{29}{7}$ into equation $(i)$:
$x(1 + 2(-\frac{29}{7})) + y(2 - \frac{29}{7}) + z(3 + \frac{29}{7}) + (5(-\frac{29}{7}) - 4) = 0$
$x(\frac{7 - 58}{7}) + y(\frac{14 - 29}{7}) + z(\frac{21 + 29}{7}) + (\frac{-145 - 28}{7}) = 0$
$-\frac{51}{7}x - \frac{15}{7}y + \frac{50}{7}z - \frac{173}{7} = 0$
Multiplying by $-7$,we get the required equation: $51x + 15y - 50z + 173 = 0$.