The shortest distance between the lines $$\dfrac{x-4}{4} = \dfrac{y+2}{5} = \dfrac{z+3}{3}$$ and $$\dfrac{x-1}{3} = \dfrac{y-3}{4} = \dfrac{z-4}{2}$$ is

We need the shortest distance between the lines:

$$L_1: \dfrac{x-4}{4} = \dfrac{y+2}{5} = \dfrac{z+3}{3}$$

$$L_2: \dfrac{x-1}{3} = \dfrac{y-3}{4} = \dfrac{z-4}{2}$$

The position vector of a point on L1 is $$\vec{a_1} = (4, -2, -3)$$ with direction vector $$\vec{d_1} = (4, 5, 3)$$, and for L2 we have $$\vec{a_2} = (1, 3, 4)$$ with direction vector $$\vec{d_2} = (3, 4, 2)$$.

The cross product of the direction vectors is $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{vmatrix} = \hat{i}(10 - 12) - \hat{j}(8 - 9) + \hat{k}(16 - 15) = (-2, 1, 1)$$.

The magnitude of this cross product is $$|\vec{d_1} \times \vec{d_2}| = \sqrt{4 + 1 + 1} = \sqrt{6}$$.

The vector connecting the points on the two lines is $$\vec{a_2} - \vec{a_1} = (1 - 4, 3 - (-2), 4 - (-3)) = (-3, 5, 7)$$.

Using the formula for the shortest distance between skew lines, we have $$d = \dfrac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$$.

Substituting gives $$d = \dfrac{|(-3)(-2) + 5 \times 1 + 7 \times 1|}{\sqrt{6}} = \dfrac{|6 + 5 + 7|}{\sqrt{6}} = \dfrac{18}{\sqrt{6}} = \dfrac{18\sqrt{6}}{6} = 3\sqrt{6}$$.

The correct answer is Option D: $$3\sqrt{6}$$.