If the sum of squares of distances of a point from the planes x + y + z = 0 , x - z = 0 and x - 2 y + z = 0 is p 2 then locus of the point is

Step-by-step solution

Let distances of a point P x , y , z from the planes x + y + z = 0 ,   x - z = 0 and x - 2 y + z = 0 are x + y + z 3 ,   x - z 2 and x - 2 y + z 6 respectively, then the sum of the squares of distances, is as ⇒ x + y + z 3 2 + x - z 2 2 + x - 2 y + z 6 2 = p 2 ⇒ 2 x + y + z 2 + 3 x - z 2 + x - 2 y + z 2 = 6 p 2 ⇒    2 x 2 + 2 y 2 + 2 z 2 + 4 x y + 4 y z + 4 z x + 3 x 2 + 3 z 2 - 6 x z + x 2 + 4 y 2 + z 2 - 4 x y - 4 y z + 2 x z = 6 p 2 ⇒ 6 x 2 + 6 y + 6 z 2 = 6 p 2 &#8658