The vertices of a triangle ABC are A = (2, 0, 2), B = (-1, 1, 1) and C = (1, - 2, 4). The point D and E divide the sides AB and CA in the ratio 1:2, respectively. Another point F is taken in space such that perpendicular drawn from F on ∆ABC meets the triangle at the point of intersection of the line segment CD and BE, say P. If the distance of F from the plane of the ∆ABC is √2 units, then
The vector PF is
(A) 7j + 7k
(B) 7/√2(j + k)
(C) j + k
(D) of these
