Let a line L passing through the point P (1,1,1) be perpendicular to the lines x-4 4 = y-1 1 = z-1 1 and x-17 1 = y-71 1 = z 0 . Let the line L intersect the y z- plane at the point Q. Another line parallel to L and passing through the point S (1,0,-1) intersects the y z -plane at the point R. Then the square of the area of the parallelogram PQRS is equal to _ _ _ _ .

Step-by-step solution

Line L passes through P(1,1,1) perpendicular to two lines with directions (4, 1, 1) and (1, 1, 0) . Direction of L: (4, 1, 1) (1, 1, 0) = (-1, 1, 3) Line L: (x, y, z) = (1, 1, 1) + t(-1, 1, 3) Q is intersection with yz-plane (x=0): Setting 1-t=0 gives t=1 , so Q = (0, 2, 4) Line through S(1,0,-1) parallel to L: (x, y, z) = (1, 0, -1) + s(-1, 1, 3) R is intersection with yz-plane: Setting 1-s=0 gives s=1 , so R = (0, 1, 2) For parallelogram PQRS: PQ = (-1, 1, 3) and PS = (0, -1, -2) Area = | PQ PS | = |(1, -2, 1)| =