The shortest distance between the lines $$\frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5}$$ and $$\frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3}$$ is

We need to find the shortest distance between two skew lines. Since $$L_1$$ passes through $$\vec{a_1} = (3, -15, 9)$$ with direction $$\vec{d_1} = (2, -7, 5)$$ and $$L_2$$ passes through $$\vec{a_2} = (-1, 1, 9)$$ with direction $$\vec{d_2} = (2, 1, -3)$$, we first compute the vector connecting their points.

Substituting the coordinates gives $$\vec{a_2} - \vec{a_1} = (-1-3, 1-(-15), 9-9) = (-4, 16, 0)$$.

Next, we compute the cross product by evaluating the determinant $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-7&5\\2&1&-3\end{vmatrix}$$. For the $$\hat{i}$$-component we have $$(-7)(-3) - (5)(1) = 21 - 5 = 16$$, for the $$\hat{j}$$-component $$-((2)(-3) - (5)(2)) = -(-6-10) = 16$$, and for the $$\hat{k}$$-component $$(2)(1) - (-7)(2) = 2 + 14 = 16$$, giving $$\vec{d_1} \times \vec{d_2} = (16, 16, 16)$$.

Then the magnitude is $$|\vec{d_1} \times \vec{d_2}| = \sqrt{16^2 + 16^2 + 16^2} = 16\sqrt{3}$$.

Finally, applying the shortest distance formula $$SD = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$$ and substituting the computed values gives $$= \frac{|(-4)(16) + (16)(16) + (0)(16)|}{16\sqrt{3}} = \frac{|-64 + 256 + 0|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$$.

The correct answer is Option 2: $$4\sqrt{3}$$.