Let the equation of the plane, that passes through the point (1, 4, -3) and contains the line of intersection of the
planes 3x - 2y + 4z - 7 = 0
and x + 5y - 2z + 9 = 0, be
αx + βy + γz + 3 = 0, then α + β + γ is equal to :

Solution

Maths sir · 5 steps

Use the family P₁ + λ P₂ = 0, fix λ with the point, then match the constant term to +3.

1. Family of planes. Every plane through the line of intersection of P₁ = 0 and P₂ = 0 has the form P₁ + λ P₂ = 0 for some number λ.

   (3x - 2y + 4z - 7) + λ(x + 5y - 2z + 9) = 0

2. Use the point. Put in (1, 4, -3). The first bracket gives 3 - 8 - 12 - 7 = -24 and the second gives 1 + 20 + 6 + 9 = 36. So -24 + 36λ = 0, which gives λ = 2/3.
3. Write the plane. Multiply through by 3: 3(3x - 2y + 4z - 7) + 2(x + 5y - 2z + 9) = 0, which is 11x + 4y + 8z - 3 = 0. Check: 11 + 16 - 24 - 3 = 0.
4. Match the form. The given form has constant +3, so multiply by -1: -11x - 4y - 8z + 3 = 0. That makes α = -11, β = -4, γ = -8.
5. So. α + β + γ = -11 - 4 - 8 = -23, which is option (a).

The idea to keep. A plane through the intersection of two planes is P₁ + λ P₂ = 0. After you find it, scale it so it matches the given form exactly before reading off coefficients.

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