The shortest distance between the lines $$\frac{x-3}{4} = \frac{y+7}{-11} = \frac{z-1}{5}$$ and $$\frac{x-5}{3} = \frac{y-9}{-6} = \frac{z+2}{1}$$ is:

Line 1: Point $$A(3,-7,1)$$, direction $$\vec{d_1} = (4,-11,5)$$.

Line 2: Point $$B(5,9,-2)$$, direction $$\vec{d_2} = (3,-6,1)$$.

$$\vec{AB} = (2, 16, -3)$$.

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1 \end{vmatrix} = (-11+30)\hat{i} - (4-15)\hat{j} + (-24+33)\hat{k} = 19\hat{i} + 11\hat{j} + 9\hat{k}$$

$$|\vec{d_1} \times \vec{d_2}| = \sqrt{361 + 121 + 81} = \sqrt{563}$$.

Shortest distance = $$\frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$$

$$= \frac{|2(19) + 16(11) + (-3)(9)|}{\sqrt{563}} = \frac{|38 + 176 - 27|}{\sqrt{563}} = \frac{187}{\sqrt{563}}$$

The correct answer is Option 2: $$\frac{187}{\sqrt{563}}$$.