Question:

Step 1: To find the shortest distance between two lines, we first need to identify a point on each line and then find the direction vectors of the lines. The equations given represent two lines in three-dimensional space, and we can rewrite them in vector form to extract the necessary information. Step 2: The first line can be represented by a point $P(3, -15, 9)$ and a direction vector $\vec{d_1} = <2, -7, 5>$. Similarly, the second line can be represented by a point $Q(-1, 1, 9)$ and a direction vector $\vec{d_2} = <2, 1, -3>$. Step 3: The shortest distance between two lines can be found using the formula $SD = \frac{|\vec{PQ} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$, where $\vec{PQ}$ is the vector from a point on the first line to a point on the second line, and $\vec{d_1} \times \vec{d_2}$ is the cross product of the direction vectors of the two lines. Step 4: First, we find $\vec{PQ} = Q - P = (-1 - 3, 1 - (-15), 9 - 9) = (-4, 16, 0)$. Then, we calculate the cross product $\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix}$. Step 5: Expanding the determinant for the cross product gives us $\vec{d_1} \times \vec{d_2} = \hat{i}((-7)(-3) - (5)(1)) - \hat{j}((2)(-3) - (5)(2)) + \hat{k}((2)(1) - (-7)(2)) = \hat{i}(21 - 5) - \hat{j}(-6 - 10) + \hat{k}(2 + 14) = 16\hat{i} + 16\hat{j} + 16\hat{k} = <16, 16, 16>$. Step 6: Now, we calculate the numerator of the formula, which is the absolute value of the dot product of $\vec{PQ}$ and $\vec{d_1} \times \vec{d_2}$. This gives us $|(-4, 16, 0) \cdot (16, 16, 16)| = |(-4)(16) + (16)(16) + (0)(16)| = |-64 + 256| = 192$. Step 7: The denominator of the formula requires the magnitude of the cross product $\vec{d_1} \times \vec{d_2}$, which is $|\vec{d_1} \times \vec{d_2}| = \sqrt{(16)^2 + (16)^2 + (16)^2} = \sqrt{256 + 256 + 256} = \sqrt{768} = 16\sqrt{3}$. Step 8: Substituting the calculated values into the formula gives $SD = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}}$. Rationalizing the denominator yields $SD = \frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}$. Step 9: Therefore, the shortest distance between the two lines is $4\sqrt{3}$, which corresponds to Option 2. The final answer is $4\sqrt{3}$, and the correct option is Option 2.