Let in a $\triangle ABC$, the length of the side $AC$ be $6$, the vertex $B$ be $(1, 2, 3)$ and the vertices $A, C$ lie on the line $\frac{x-3}{2} = \frac{y-7}{2} = \frac{z-7}{2}$. Then the area (in sq. units) of $\triangle ABC$ is