If Q(0, -1, -3) is the image of the point P in the plane 3x - y + 4z = 2 and R is the point (3, -1, -2), then the area (in sq. units) of ΔPQR is:

Correct option is D. 912
R lies on the plane.
DQ = |1 - 1; 2 - 2| = √9 + 1 + 16 = √26
⇒ PQ = 2√26
Now, RQ = √9 + 1 = √10 ⇒ RD = √10 - 1; 3/2 = 7/2
Hence, ar(ΔPQR) = 1/2 × 2√26 × 7/2 = 912.