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 plane is

a(x + 1) + b(y – 3) + c(z – 2) = 0

⇒ ax + by + cz = 3b + 2c – a …… (1)

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

x + 2y + 3z = 5 and 3x + 3y + z = 0

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 + 3c = 0 …… (b)

3a + 3b + c = 0 …… (c)

Solving (b) and (c) we get,

∴a = – 7λ, b = 8λ, c = – 3λ

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

(– 7λ)x + (8λ)y + (– 3λ)z = 3(8λ) + 2(– 3λ) + 7λ

⇒ – 7λx + 8λy – 3λz = 24λ – 6λ + 7λ

⇒ – 7λx + 8λy – 3λz = 25λ

Dividing both sides by (– λ) we get

7x – 8y + 3z – 25 = 0

So, the equation of required planes is 7x – 8y + 3z – 25 = 0