Let P be the plane, which contains the line of intersection of the planes, x + y + z - 6 = 0 and 2 x + 3 y + z + 5 = 0 and it is perpendicular to the x y -plane. Then the distance of the point 0 , 0 , 256 from P is equal to:

Step-by-step solution

Using family of planes, equation to plane passing through the line of intersection of the given planes, is λ x + y + z - 6 + 2 x + 3 y + z + 5 = 0 ⇒ λ + 2 x + λ + 3 y + λ + 1 z + 5 - 6 λ = 0 If a plane is perpendicular to x y plane, then it doesn't contain the term of z . Hence, λ + 1 = 0 ⇒ λ = - 1 . Thus, the plane is x + 2 y + 11 = 0 The length of perpendicular from a point x 1 ,   y 1 ,   z 1 to the plane a x + b y + c z + d = 0 is a x 1 + b y 1 + c z