JEE Main 2012 Mathematics question
Consider the following planes
P : x + y - 2z + 7 = 0
Q : x + y + 2z + 2 = 0
R : 3x + 3y - 6z - 11 = 0
Answer: (4)
Solution
Normals: P (1, 1, −2), Q (1, 1, 2), R (3, 3, −6) = 3 × P's normal. So P ∥ R, option (4). (Q and R: 3 + 3 − 12 ≠ 0, not perpendicular.)