The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes
3x + y - 2z = 5 and 2x - 5y - z = 7, is :

Solution

Maths sir · 3 steps

Perpendicular to both planes means your normal is perpendicular to both normals, so take their cross product.

1. Find the normal. If two planes are perpendicular, their normals are perpendicular. The required plane's normal must be perpendicular to both (3, 1, -2) and (2, -5, -1), and their cross product gives exactly such a vector.

   n⃗ = |î, ĵ, k̂; 3, 1, -2; 2, -5, -1| = -11î - ĵ - 17k̂

2. Write the plane. Take the normal as (11, 1, 17) and pass the plane through (1, 2, -3): 11(x-1) + 1(y-2) + 17(z+3) = 0. Expanding gives 11x + y + 17z + 38 = 0.

   Option	Through (1, 2, -3)?	n⃗ · (3, 1, -2)	n⃗ · (2, -5, -1)
   (a) (6, -5, 2)	yes	9	35
   (b) (3, -10, -2)	yes	3	58
   (c) (6, -5, -2)	yes	17	39
   (d) (11, 1, 17)	yes	0	0

3. So. All four options pass through the point, but only (d) has a normal with zero dot product with both given normals. The answer is (d) 11x + y + 17z + 38 = 0.

The idea to keep. A plane perpendicular to two planes has its normal along the cross product of their normals. Checking that a point lies on the plane is not enough. Always check the normal too.

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