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:

In triangle ABC, we have AC = 6, B = (1,2,3), and points A and C lie on the line $$\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$$. Parametrizing this line by $$(6+3t, 7+2t, 7-2t)$$ gives all points on AC.

To compute the altitude from B to line AC, choose the point corresponding to $$t=0$$, namely $$P=(6,7,7)$$, on the line, and note that its direction vector is $$\vec{d} = (3,2,-2)$$. The vector from $$P$$ to $$B$$ is $$\vec{PB} = (1-6, 2-7, 3-7) = (-5,-5,-4)\,.$$

The area of the parallelogram determined by $$\vec{PB}$$ and $$\vec{d}$$ is the magnitude of their cross product:

$$\vec{PB} \times \vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-5&-5&-4\\3&2&-2\end{vmatrix} = \hat{i}(10+8) - \hat{j}(10+12) + \hat{k}(-10+15) = (18, -22, 5)\,,$$

so $$|\vec{PB} \times \vec{d}| = \sqrt{324+484+25} = \sqrt{833}$$. Since $$|\vec{d}| = \sqrt{9+4+4} = \sqrt{17}\,,$$ the distance from B to the line is

$$\frac{\sqrt{833}}{\sqrt{17}} = \sqrt{\frac{833}{17}} = \sqrt{49} = 7\,,$$

which is the length of the altitude.

Finally, the area of triangle ABC is half the product of its base AC and its height:

$$\text{Area} = \frac{1}{2} \times AC \times h = \frac{1}{2} \times 6 \times 7 = 21\,.$$

The correct answer is Option 2: 21.