Let L be the line of intersection of the planes 2x - y + 1 = 0 and 3x - z - 1 = 0 . Let P be the image of the point Q (2, 4, 1) in the line L . If R is the point (3, 1, 4) and the area of the triangle PQR is , then the value of ^2 is :

Step-by-step solution

The line L is the intersection of the planes P₁: 2x - y + 1 = 0 and P₂: 3x - z - 1 = 0 . The direction vector of L is d = n₁ n₂ = (2 i - j ) (3 i - k ) = i + 2 j + 3 k . To find a point on L , let x = 0 . Then y = 1 and z = -1 . The point is A(0, 1, -1) . The equation of line L is x 1 = y-1 2 = z+1 3 = t . Let S(t, 2t+1, 3t-1) be the foot of the perpendicular from Q(2, 4, 1) to L . The vector SQ = (2-t) i + (4-(2t+1)) j + (1-(3t-1)) k = (2-t) i + (3-2t) j + (2-3t) k . Since SQ is perpendicular to d , SQ d = 0 : 1(2