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 $\dfrac{x - 6}{3} = \dfrac{y - 7}{2} = \dfrac{z - 7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is: