Let the plane P contain the line 2 x + y - z - 3 = 0 = 5 x - 3 y + 4 z + 9 and be parallel to the line x + 2 2 = 3 - y - 4 = z - 7 5 . Then the distance of the point A 8 , - 1 , - 19 from the plane P measured parallel to the line x - 3 = y - 5 4 = 2 - z - 12 is equal to _________.

Step-by-step solution

Let the Plane containing the line 2 x + y - 3 - 3 = 0 = 5 x - 3 y + 4 z + 9 is given by, ⇒   2 x + y - z - 3 + λ 5 x - 3 y + 4 z + 9 = 0 ⇒ x 2 + 5 λ + y 1 - 3 λ + z 4 λ - 1 + 9 λ - 3 = 0 Also, this plane is parallel to the line x + 2 2 = 3 - y - 4 = z - 7 5 ⇒ x + 2 2 = y - 3 4 = z - 7 5 Now normal of the plane will be perpendicular to the line, so using perpendicular condition we get, ∴   2 + 5 λ 2 + 1 - 3 λ 4 + 4 λ - 1 5 = 0 ⇒ 4 + 10 &