Given planes, 3x – 6y – 2z = 7 and 2x + y – λz = 5

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)

We have, a₁ = 3, b₁ = – 6, c₁ = – 2 and a₂ = 2, b₂ = 1, c₂ = – λ

Using (a) we have,

a₁a₂ + b₁b₂ + c₁c₂ = (3)(2) + (– 6)(1) + (– 2)(– λ) = 0

⇒ 6 – 6 + 2λ = 0

⇒ 0 = – 2λ

⇒ 0 = λ

For λ = 0 the given planes are perpendicular to each other.