Let $$O$$ be the origin, and $$M$$ and $$N$$ be the points on the lines $$\frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3}$$ and $$\frac{x+8}{12} = \frac{y+2}{5} = \frac{z+11}{9}$$ respectively such that $$MN$$ is the shortest distance between the given lines. Then $$\vec{OM} \cdot \vec{ON}$$ is equal to ______.

We need to find $$\vec{OM} \cdot \vec{ON}$$ where $$M$$ and $$N$$ are points on the given lines such that $$MN$$ is the shortest distance between them.

Parametrize the lines: line $$L_1$$: $$\frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3} = t$$ gives $$M = (5+4t, \, 4+t, \, 5+3t)$$.

Line $$L_2$$: $$\frac{x+8}{12} = \frac{y+2}{5} = \frac{z+11}{9} = s$$ gives $$N = (-8+12s, \, -2+5s, \, -11+9s)$$.

Direction vectors: $$\vec{d_1} = (4, 1, 3)$$, $$\vec{d_2} = (12, 5, 9)$$.

Find $$\vec{d_1} \times \vec{d_2}$$ (direction of shortest distance): $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 3 \\ 12 & 5 & 9 \end{vmatrix} = (9-15)\hat{i} - (36-36)\hat{j} + (20-12)\hat{k} = (-6, 0, 8)$$

Set up equations for shortest distance: $$\vec{MN}$$ must be parallel to $$(-6, 0, 8)$$, so $$\vec{MN} \cdot \vec{d_1} = 0$$ and $$\vec{MN} \cdot \vec{d_2} = 0$$.

$$\vec{MN} = (-13+12s-4t, \, -6+5s-t, \, -16+9s-3t)$$

From $$\vec{MN} \cdot \vec{d_1} = 0$$:

$$4(-13+12s-4t) + 1(-6+5s-t) + 3(-16+9s-3t) = 0$$

$$-52+48s-16t-6+5s-t-48+27s-9t = 0$$

$$80s - 26t - 106 = 0 \quad \cdots(1)$$

From $$\vec{MN} \cdot \vec{d_2} = 0$$:

$$12(-13+12s-4t) + 5(-6+5s-t) + 9(-16+9s-3t) = 0$$

$$-156+144s-48t-30+25s-5t-144+81s-27t = 0$$

$$250s - 80t - 330 = 0 \quad \cdots(2)$$

From (1): $$80s - 26t = 106$$, i.e., $$40s - 13t = 53$$.

From (2): $$250s - 80t = 330$$, i.e., $$25s - 8t = 33$$.

Solving: From the first, $$t = \frac{40s - 53}{13}$$. Substituting:

$$25s - 8 \cdot \frac{40s - 53}{13} = 33$$

$$325s - 320s + 424 = 429$$

$$5s = 5 \implies s = 1$$

$$t = \frac{40 - 53}{13} = \frac{-13}{13} = -1$$

Find M and N: $$M = (5-4, \, 4-1, \, 5-3) = (1, 3, 2)$$

$$N = (-8+12, \, -2+5, \, -11+9) = (4, 3, -2)$$

Compute $$\vec{OM} \cdot \vec{ON}$$: $$ \vec{OM} \cdot \vec{ON} = (1)(4) + (3)(3) + (2)(-2) = 4 + 9 - 4 = 9 $$

The answer is $$\boxed{9}$$.