Consider a $$\triangle ABC$$ where $$A(1, 3, 2)$$, $$B(-2, 8, 0)$$ and $$C(3, 6, 7)$$. If the angle bisector of $$\angle BAC$$ meets the line BC at D, then the length of the projection of the vector $$\vec{AD}$$ on the vector $$\vec{AC}$$ is:

We have $$A(1, 3, 2)$$, $$B(-2, 8, 0)$$, $$C(3, 6, 7)$$.

First, find $$AB$$ and $$AC$$:

$$\vec{AB} = (-3, 5, -2)$$, so $$AB = \sqrt{9 + 25 + 4} = \sqrt{38}$$

$$\vec{AC} = (2, 3, 5)$$, so $$AC = \sqrt{4 + 9 + 25} = \sqrt{38}$$

Since $$AB = AC$$, the triangle is isosceles. By the angle bisector theorem, D divides BC in the ratio $$AB : AC = 1 : 1$$, so D is the midpoint of BC.

$$D = \left(\frac{-2+3}{2}, \frac{8+6}{2}, \frac{0+7}{2}\right) = \left(\frac{1}{2}, 7, \frac{7}{2}\right)$$

$$\vec{AD} = \left(\frac{1}{2} - 1, 7 - 3, \frac{7}{2} - 2\right) = \left(-\frac{1}{2}, 4, \frac{3}{2}\right)$$

The projection of $$\vec{AD}$$ on $$\vec{AC}$$ is given by:

$$\text{Projection} = \frac{\vec{AD} \cdot \vec{AC}}{|\vec{AC}|}$$

$$\vec{AD} \cdot \vec{AC} = \left(-\frac{1}{2}\right)(2) + (4)(3) + \left(\frac{3}{2}\right)(5) = -1 + 12 + \frac{15}{2} = 11 + \frac{15}{2} = \frac{37}{2}$$

$$|\vec{AC}| = \sqrt{38}$$

$$\text{Projection} = \frac{37/2}{\sqrt{38}} = \frac{37}{2\sqrt{38}}$$

The answer is $$\boxed{\dfrac{37}{2\sqrt{38}}}$$, which corresponds to Option (1).