Step 1:

Let the equation of a plane passing through the line of intersection of the planes $(x+2y+3z-4)=0$ and $(2x+y-z+5)=0$ be

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

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

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

$(a_1a_2+b_1b_2+c_1c_2)=0$

Therefore $5(1+2 \lambda)+3 (2+\lambda)+6(3-\lambda)=0$

On simplifying we get,

$5+10 \lambda+6+3 \lambda+18-6 \lambda=0$

$=>7 \lambda+29=0$

Therefore $ \lambda=\large\frac{-29}{7}$

Step 2:

Substituting for $\lambda$ in equ (1) we get,

$x\bigg[1+2 .\bigg(\large\frac{-29}{7}$$\bigg)\bigg]+y\bigg[2 +\bigg(\large\frac{-29}{7}$$\bigg)\bigg]+z\bigg[3- \bigg(\large\frac{-29}{7}$$\bigg)\bigg]-4+5\bigg(\large\frac{-29}{7}\bigg)=0$

On simplifying we get,

$\large\frac{-51}{7}x-\frac{15}{7} y+\frac{50}{7} z-\frac{173}{7}=0$

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

This is the required solution of the plane.