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