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The given lines are: Line 1: $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ Line 2: $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ For Line 1, the direction vector $\vec{b_1} = 2\hat{i} + \hat{j} - 3\hat{k}$ and a point on the line is $A(-1, 1, 9)$. For Line 2, the direction vector $\vec{b_2} = 2\hat{i} - 7\hat{j} + 5\hat{k}$ and a point on the line is $B(3, -15, 9)$.
The vector connecting the points $A$ and $B$ is: $\vec{AB} = (3 - (-1))\hat{i} + (-15 - 1)\hat{j} + (9 - 9)\hat{k} = 4\hat{i} - 16\hat{j} + 0\hat{k}$
The cross product of the direction vectors $\vec{b_1}$ and $\vec{b_2}$ is: $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 2 & -7 & 5 \end{vmatrix} = (5 - 21)\hat{i} - (10 - (-6))\hat{j} + (-14 - 2)\hat{k} = -16\hat{i} - 16\hat{j} - 16\hat{k}$
The magnitude of the cross product is: $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-16)^2 + (-16)^2 + (-16)^2} = \sqrt{3 \cdot 16^2} = 16\sqrt{3}$
The scalar triple product is the dot product of $\vec{AB}$ and $\vec{b_1} \times \vec{b_2}$: $\vec{AB} \cdot (\vec{b_1} \times \vec{b_2}) = (4\hat{i} - 16\hat{j} + 0\hat{k}) \cdot (-16\hat{i} - 16\hat{j} - 16\hat{k}) = (4)(-16) + (-16)(-16) + (0)(-16) = -64 + 256 + 0 = 192$
The shortest distance $d$ between the lines is given by: $d = \frac{|\vec{AB} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} = \frac{|192|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}$
Final Answer: $4\sqrt{3}$