The vertices B and C of a triangle ABC lie on the line $$\frac{x}{1}=\frac{1-y}{-2}=\frac{z-2}{3}$$ The coordinates of A and B are (1, 6, 3) and (4, 9, $$\alpha$$) respectively and C is at a distance of 10 units from B. The area (in sq. units) of $$\triangle$$ABC is :

A

5$$\sqrt{13}$$

B

10$$\sqrt{13}$$

C

15$$\sqrt{13}$$

D

20$$\sqrt{13}$$