We know that solution of a plane passing through (x₁,y₁,z₁) is given as –

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

The required plane passes through (1, – 3, – 2), so the equation of the plane is

a(x – 1) + b(y + 3) + c(z + 2) = 0

⇒ ax + by + cz = a – 3b – 2c …… (1)

Now, the required plane is also perpendicular to the planes,

x + 2y + 2z = 5 and 3x + 3y + 2z = 8

We know that planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are at right angles if,

a₁a₂ + b₁b₂ + c₁c₂ = 0 …… (a)

Using (a) we have,

a + 2b + 2c = 0 …… (b)

3a + 3b + 2c = 0 …… (c)

Solving (b) and (c) we get,

∴a = – 2λ, b = 4λ, c = – 3λ

Putting values of a,b,c in equation (1) we get,

(– 2λ)x + (4λ)y + (– 3λ)z = (– 2)λ – 3(4λ) – 2(– 3λ)

⇒ – 2λx + 4λy – 3λz = – 2λ – 12λ + 6λ

⇒ – 2λx + 4λy – 3λz = – 8λ

Dividing both sides by (– λ) we get

2x – 4y + 3z = 8

So, the equation of the required planes is 2x – 4y + 3z = 8