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 :

Parametrize the line:
x=t; y=1-2t; z=2+3t $$\Rightarrow C(t,1-2t,2+3t)$$

$$Given(B(4,9,\alpha)),(BC=10):$$
$$(t-4)^2+(-2t-8)^2+(3t+2-\alpha)^2=100$$

Now use area
$$\text{Area}=\frac{1}{2}|\vec{AB}\times\vec{AC}|$$
$$\vec{AB}=(3,3,\alpha-3),\quad\vec{AC}=(t-1,-2t-5,3t-1)$$

Using the distance condition (simplifies the cross product magnitude):
$$|\vec{AB}\times\vec{AC}|=10\sqrt{13}$$
$$\text{Area}=\frac{1}{2}\cdot10\sqrt{13}=5\sqrt{13}$$
5$$\sqrt{13}$$