A plane passes through (1,-2,1) and is perpendicular to the planes 2 x-2 y + z =0 and x- y +2 z =4 . The distance of the point (1,2,2) from this plane is _______ units.

Step-by-step solution

Find the plane perpendicular to both given planes and passing through (1, -2, 1) . The required plane's normal vector n is perpendicular to normals n₁ = (2, -2, 1) and n₂ = (1, -1, 2) , so n = n₁ n₂ = ( -3, -3, 0 ) , which simplifies to (1, 1, 0) . Using point-normal form with (1, -2, 1) : 1(x - 1) + 1(y + 2) + 0(z - 1) = 0 simplifies to x + y + 1 = 0 . Compute the distance from point (1, 2, 2) to the plane. The distance formula gives d = |1 1 + 1 2 + 0 2 + 1| 1^2 + 1^2 + 0^2 = 4 2 = 2 2 . 2 2 corresponds to option