Let the plane containing the line of intersection of the planes
P₁ : x+(λ+4)y+z=1 and
P₂ : 2x+y+z=2
pass through the points (0, 1, 0) and (1, 0, 1). Then the distance of
the point (2λ,λ,-λ) from the plane P₂ is :

Solution

Maths sir · 5 steps

Write the family P₁ + μ P₂ = 0, then let the two given points fix μ and λ.

1. Family of planes. Every plane through the line where P₁ and P₂ meet can be written as P₁ + μ P₂ = 0 for some number μ.

   x + (λ+4)y + z - 1 + μ (2x + y + z - 2) = 0

2. Use (1, 0, 1). Put x = 1, y = 0, z = 1: (1 + 0 + 1 - 1) + μ(2 + 0 + 1 - 2) = 0, so 1 + μ = 0 and μ = -1. This point does not involve λ, so use it first.
3. Use (0, 1, 0). Put x = 0, y = 1, z = 0: (λ + 4 - 1) + μ(1 - 2) = 0, so λ + 3 - μ = 0. With μ = -1 this gives λ = -4.
4. Distance from P2. The point (2λ, λ, -λ) is (-8, -4, 4). Its distance from 2x + y + z - 2 = 0 is (|2(-8) + (-4) + 4 - 2|)/√(2² + 1² + 1²) = 18/√6.
5. So. 18/√6 = 18√6/6 = 3√6, which is option (b).

The idea to keep. For a plane through the line of two planes, always start from P₁ + μ P₂ = 0. Each point it passes through gives you one equation, so two points can fix two unknowns.

Answer from the NTA final answer key. Solution by PrepWiser.