Let a line L₁ be given by the intersection of the planes x + y - z = 0 and 2x - y + z = 3 . Let another line L₂ be given by the intersection of the planes x - 2y + z = 0 and 3x + y - 2z = k . If the lines L₁ and L₂ intersect, and d is the distance of their point of intersection from the point Q(4, 6, 3) , then the value of d^2 + k is :

Step-by-step solution

Since the lines L₁ and L₂ intersect, their point of intersection must satisfy the equations of all four planes simultaneously. The equations are: x + y - z = 0 2x - y + z = 3 x - 2y + z = 0 3x + y - 2z = k Adding the first two equations: (x + y - z) + (2x - y + z) = 0 + 3 3x = 3 x = 1 . Substitute x = 1 into the first and third equations: 1 + y - z = 0 y - z = -1 1 - 2y + z = 0 -2y + z = -1 Adding these two new equations: (y - z) + (-2y + z) = -1 - 1 -y = -2 y = 2 . Substituting y = 2 into y - z = -1 gives 2 - z =