Let the plane P pass through the intersection of the planes 2x+3y-z=2 and x+2y+3z=6, and be perpendicular to the plane 2x+y-z+1=0. If d is the distance of P from the point (-7, 1, 1), then d² is equal to :

Solution

Maths sir · 4 steps

Write the family of planes through the line, then let the perpendicular condition fix λ.

1. Family of planes. Every plane through the line of intersection has the form (2x+3y-z-2) + λ(x+2y+3z-6) = 0. Its normal is (2+λ, 3+2λ, -1+3λ).
2. Use perpendicularity. Two planes are perpendicular when their normals have zero dot product. Dot this normal with (2, 1, -1): 2(2+λ) + (3+2λ) - (-1+3λ) = 8 + λ = 0, so λ = -8.
3. Find the plane. Put λ = -8 into the family: -6x - 13y - 25z + 46 = 0. Multiply by -1 to get P: 6x + 13y + 25z - 46 = 0.

   d = (|6(-7) + 13(1) + 25(1) - 46|)/√(6²+13²+25²) = (|-50|)/√830 = 50/√830

4. So. d² = 2500/830 = 250/83, which is option (a).

The idea to keep. Plane through the intersection of two planes: always start with P₁ + λ P₂ = 0. One extra condition, such as passing through a point or being perpendicular to a plane, fixes λ.

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