The equation of the plane through the intersection of

the planes x + 2y + 3z=4 or, x + 2y + 3z - 4=0 and

2x + y–z=–5 or, 2x + y–z + 5=0 is given as,

(x + 2y + 3z - 4) + λ(2x + y–z + 5)=0

[where λ is a scalar]

x(1 + 2λ) + y(2 + λ) + z(3 - λ) - 4 + 5λ=0

Given, that the required plane is perpendicular to the plane 5x + 3y + 6z + 8=0 so, we should have,

5(1 + 2λ) + 3(2 + λ) + 6(3 - λ)=0

5 + 10λ + 6 + 3λ + 18 - 6λ=0

29 + 7λ=0

Therefore, the equation of the required plane is,

7(x + 2y + 3z - 4) - 29(2x + y–z + 5)=0

7x + 14y + 21z - 28 - 58x - 29y + 29z - 145=0

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

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