Given two lines are represented as:

$ \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} $

The formula for the shortest distance between two lines is:

$ d = \frac{|\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right)|}{|\vec{b}_1 \times \vec{b}_2|} $

From the given lines:

$ \vec{a}_1 = (3, -15, 9) $

$ \vec{a}_2 = (-1, 1, 9) $

$ \vec{b}_1 = (2, -7, 5) $

$ \vec{b}_2 = (2, 1, -3) $

Calculate the difference:

$ \vec{a}_2 - \vec{a}_1 = (-1-3, 1+15, 9-9) = (-4, 16, 0) $

Next, compute the cross product:

$ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = (16\hat{i} + 16\hat{j} + 16\hat{k}) $

The magnitude of the cross product is:

$ |\vec{b}_1 \times \vec{b}_2| = |(16\hat{i} + 16\hat{j} + 16\hat{k})| = 16\sqrt{3} $

Calculate the dot product:

$ (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-4\hat{i} + 16\hat{j} + 0\hat{k}) \cdot (16\hat{i} + 16\hat{j} + 16\hat{k}) = -64 + 256 + 0 = 192 $

Finally, find the shortest distance:

$ d = \frac{|192|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = 4\sqrt{3} $

Thus, the shortest distance is:

$ \boxed{4 \sqrt{3}} $