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

Step-by-step solution

A plane P pass through the intersection of planes P 1 and P 2 can be written as, P ≡ P 1 + λ P 2 = 0 2 + λ x + 3 + 2 λ y + 3 λ - 1 z - 2 + 6 λ = 0 Given, P ⊥ P 3 ( 2 x + y - z + 1 = 0 ) ∴   n → . n → 3 = 0 ⇒ 2 λ + 2 + 2 λ + 3 - 3 λ - 1 = 0 ⇒ 4 + 2 λ + 2 λ + 3 - 3 λ + 1 = 0 ⇒ λ = - 8 ∴ P ≡ - 6 x - 13 y - 25 z + 46 = 0 6 x + 13 y + 25 z - 46 = 0 Since, the distance d from ( x 0 , y 0 , z 0 ) to