We are given that,

A plane is perpendicular to another plane 5x + 3y + 6z + 8 = 0, and also contains line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.

We need to find the equation of such plane.

We know that,

The equation of a plane through the line of intersection of the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

(a₁x + b₁y + c₁z + d₁) + λ(a₂x + b₂y + c₂z + d₂) = 0

Similarly, 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) + λ(2x + y – z + 5) = 0

⇒ x + 2y + 3z – 4 + 2λx + λy – λz + 5λ = 0

⇒ x + 2λx + 2y + λy + 3z – λz – 4 + 5λ = 0

⇒ (1 + 2λ)x + (2 + λ)y + (3 – λ)z – 4 + 5λ = 0 …(i)

So, direction ratio of plane in (i) is,

Direction ratio = (1 + 2λ, 2 + λ, 3 – λ)

Since, the plane in (i) is perpendicular to the plane 5x + 3y + 6z + 8 = 0.

Then, replace x, y, z by (1 + 2λ), (2 + λ), (3 – λ) respectively in plane 5x + 3y + 6z + 8 = 0 (neglecting 8) and equate to 0. We get

5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0

⇒ 5 + 10λ + 6 + 3λ + 18 – 6λ = 0

⇒ 10λ + 3λ – 6λ + 5 + 6 + 18 = 0

⇒ 7λ + 29 = 0

⇒ 7λ = -29

Putting this value of λ in equation (i), we get

⇒ -51x – 15y + 50z – 173 = 0

⇒ 51x + 15y – 50z + 173 = 0

Thus, equation of the required plane is 51x + 15y – 50z + 173 = 0.