Given planes, 2x – 4y + 3z = 5 and x + 2y + 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 = 2, b₁ = – 4, c₁ = 3 and a₂ = 1, b₂ = 2, c₂ = λ

Using (a) we have,

a₁a₂ + b₁b₂ + c₁c₂ = (2)(1) + (– 4)(2) + (3)(λ) = 0

⇒ 2 – 8 + 3λ = 0

⇒ 6 = 3λ

⇒ 2 = λ

Hence, for λ = 2 the given planes are perpendicular.