If the equation of the plane passing through the line of intersection of the planes 2 x-y+z=3,4 x-3 y+5 z+9=0 and parallel to the line (x+1)/-2=(y+3)/4=(z-2)/5 is a x+b y+c z+6=0, then a+b+c is equal to :

Solution

Maths sir · 4 steps

Write the family of planes through the line, then make its normal perpendicular to the given line.

1. Family of planes. Every plane through the line of intersection has the form (2x - y + z - 3) + λ(4x - 3y + 5z + 9) = 0. Its normal is (2 + 4λ, -1 - 3λ, 1 + 5λ).
2. Parallel condition. A plane is parallel to a line when its normal is perpendicular to the line's direction (-2, 4, 5). So their dot product must be zero.

   -2(2 + 4λ) + 4(-1 - 3λ) + 5(1 + 5λ) = 5λ - 3 = 0 ⇒ λ = 3/5

3. Build the plane. Put λ = 3/5 in and multiply through by 5: 22x - 14y + 20z + 12 = 0. Divide by 2 so the constant is 6: 11x - 7y + 10z + 6 = 0.
4. So. a = 11, b = -7, c = 10, and a + b + c = 11 - 7 + 10 = 14, option (c).

The idea to keep. For a plane through the intersection of two planes, use P₁ + λ P₂ = 0. Parallel to a line means the normal dotted with the line's direction is zero. Scale at the end to match the given constant.

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