Let a plane P₁ contain the line L₁ : x - 3 1 = y - 2 -1 = z - 1 0 and be parallel to the line L₂ : x 1 = y 1 = z 4 . Let A be the foot of the perpendicular drawn from the point P(1, 1, 4) to the plane P₁ . If a line L₃ passes through the point A and is parallel to the line of intersection of the planes x - y + z = 1 and 2x + y - z = 2 , then the square of the shortest distance between the line L₃ and the line L₄ : x

Step-by-step solution

First, we find the equation of the plane P₁ . The normal vector n₁ to the plane P₁ is perpendicular to the direction vectors of both L₁ and L₂ . n₁ = (1, -1, 0) (1, 1, 4) = (-4, -4, 2) We can take the normal vector as (2, 2, -1) . Since P₁ contains L₁ , it passes through the point (3, 2, 1) . The equation of P₁ is 2(x - 3) + 2(y - 2) - 1(z - 1) = 0 2x + 2y - z - 9 = 0 . Next, we find the coordinates of A , the foot of the perpendicular from P(1, 1, 4) to P₁ . Let the coordinates of A be (x, y, z) . Then, x - 1 2 =