The vertices B and C of a triangle ABC lie on the line x/1=(1-y)/-2=(z -2)/3. The coordinates of A and B are (1,6,3) and (4,9, α) respectively and C…

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