The shortest distance between lines $$L_1$$ and $$L_2$$, where $$L_1: \frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+4}{2}$$ and $$L_2$$ is the line passing through the points $$A(-4, 4, 3)$$, $$B(-1, 6, 3)$$ and perpendicular to the line $$\frac{x-3}{-2} = \frac{y}{3} = \frac{z-1}{1}$$, is

The shortest distance between two skew lines $$L_1$$ and $$L_2$$ is given by the formula:

$$\text{Distance} = \frac{|(\overrightarrow{P_1P_2}) \cdot (\overrightarrow{d_1} \times \overrightarrow{d_2})|}{|\overrightarrow{d_1} \times \overrightarrow{d_2}|}$$

where $$P_1$$ is a point on $$L_1$$, $$P_2$$ is a point on $$L_2$$, $$\overrightarrow{d_1}$$ is the direction vector of $$L_1$$, and $$\overrightarrow{d_2}$$ is the direction vector of $$L_2$$.

For line $$L_1$$: $$\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+4}{2}$$

A point on $$L_1$$ is $$P_1(1, -1, -4)$$, and its direction vector is $$\overrightarrow{d_1} = (2, -3, 2)$$.

For line $$L_2$$, it passes through points $$A(-4, 4, 3)$$ and $$B(-1, 6, 3)$$. The direction vector $$\overrightarrow{d_2}$$ is:

$$\overrightarrow{d_2} = \overrightarrow{AB} = (-1 - (-4), 6 - 4, 3 - 3) = (3, 2, 0)$$

We can take $$P_2$$ as point $$A(-4, 4, 3)$$.

The vector $$\overrightarrow{P_1P_2}$$ is:

$$\overrightarrow{P_1P_2} = (-4 - 1, 4 - (-1), 3 - (-4)) = (-5, 5, 7)$$

Now, compute the cross product $$\overrightarrow{d_1} \times \overrightarrow{d_2}$$:

$$\overrightarrow{d_1} \times \overrightarrow{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{vmatrix}$$

Expanding the determinant:

$$\mathbf{i}((-3)(0) - (2)(2)) - \mathbf{j}((2)(0) - (2)(3)) + \mathbf{k}((2)(2) - (-3)(3))$$

$$= \mathbf{i}(0 - 4) - \mathbf{j}(0 - 6) + \mathbf{k}(4 - (-9))$$

$$= -4\mathbf{i} - (-6)\mathbf{j} + 13\mathbf{k}$$

$$= (-4, 6, 13)$$

The magnitude of $$\overrightarrow{d_1} \times \overrightarrow{d_2}$$ is:

$$|\overrightarrow{d_1} \times \overrightarrow{d_2}| = \sqrt{(-4)^2 + 6^2 + 13^2} = \sqrt{16 + 36 + 169} = \sqrt{221}$$

The dot product $$(\overrightarrow{P_1P_2}) \cdot (\overrightarrow{d_1} \times \overrightarrow{d_2})$$ is:

$$(-5)(-4) + (5)(6) + (7)(13) = 20 + 30 + 91 = 141$$

Therefore, the shortest distance is:

$$\frac{|141|}{\sqrt{221}} = \frac{141}{\sqrt{221}}$$

This matches option C.