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 $2x+y -z+5= 0$, is

$51x+15y-50z +173=0$

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 given by

$(x+2y+3z-4)+λ (2x+y-z+5)=0$

$⇒x (1+2λ) + y (2+λ)+z (3-λ)-4+5λ=0$

This is perpendicular to the plane $5x + 3y+6z+8=0.$

$∴5 (1+2λ)+3(2+λ)+6 (3-λ)=0$

$⇒7λ+29=0⇒λ=-\frac{29}{7}$

Putting $λ=-\frac{29}{7}$ in (i), we obtain the equation of the required plane as

$-51x-15y+50z-173=0 ⇒51x+15y-50z + 173 =0$