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 $2x + y - z + 5 = 0$.

The equation of a plane through the line of intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ is $(x + 2y + 3z - 4) + \lambda (2x + y - z + 5) = 0$

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

…..(i)
Also, this is perpendicular to the plane $5x + 3y + 6z + 8 = 0$. $\therefore 5(1 + 2\lambda ) + 3(2 + \lambda ) + 6(3 - \lambda ) = 0$

$\Rightarrow$ $5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0$
$\therefore \lambda = - 29/7$

From Eq. (i), $x\left[ {1 + 2\left( {\frac{{ - 29}}{7}} \right)} \right] + y\left( {2 - \frac{{29}}{7}} \right) + z\left( {\frac{{29}}{7} + 3} \right) - 4 + 5\left( {\frac{{ - 29}}{7}} \right) = 0$

$\Rightarrow$ $x(7 - 58) + y(14 - 29) + z(29 + 21) - 28 - 145 = 0$

$\Rightarrow$ $- 51x - 15y + 50z - 173 = 0$
So, the required equation of plane is $51x + 15y - 50z + 173 = 0$.