We know that the equation of plane passing through (x₁,y₁,z₁) is given by

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 ……(1)

So, equation of plane passing through (– 1,2,0) is

a(x + 1) + b(y – 2) + c(z – 0) = 0 ……(2)

It also passes through (2,2, – 1)

So, equation (2) must satisfy the point (2,2, – 1)

∴ a(2 + 1) + b(2 – 2) + c(– 1) = 0

⇒ 3a – c = 0 ……(3)

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 ……(4)

Here, the plane is parallel to line,

So,

a×1 + b×2 + c×1 = 0

⇒ a + 2b + c = 0 ……(5)

Solving equation (3) and (5) by cross multiplication we have,

⇒

⇒

∴ a = k, b = – 2k and c = 3k

Putting the value in equation (2)

k(x + 1) – 2k(y – 2) + 3k(z – 0) = 0

kx + k – 2ky + 4k + 3kz = 0

Dividing by k we have

x – 2y + 3z + 5 = 0

The required equation is x – 2y + 3z + 5 = 0