If the lines L₁: x-2 1 = y-1 2 = z+1 1 and L₂: x- 3 = y+1 1 = z-1 -1 intersect at a point P , and a plane passes through the point (1, , 4) and is perpendicular to L₁ , then the square of the distance of the point P from the plane is :

Step-by-step solution

Let the point of intersection be P . Since P lies on both lines, we can equate their general coordinates. For L₁ , a general point is ( + 2, 2 + 1, - 1) . For L₂ , a general point is (3 + , - 1, - + 1) . Equating the y and z coordinates: 2 + 1 = - 1 2 - = -2 - 1 = - + 1 + = 2 Adding the two equations gives 3 = 0 = 0 . Substituting = 0 into + = 2 gives = 2 . Thus, the point of intersection P is (0 + 2, 0 + 1, 0 - 1) = (2, 1, -1) . Equating the x coordinates to find : + 2 = 3 + 0 + 2 = 3(2) + = -4 . The plane passes