This problem asks us to find the shortest distance between two lines in three-dimensional space. These lines are neither parallel nor intersecting, which means they are skew lines.

1. Key Concepts and Formulas

• Skew Lines: Lines in 3D space that are not parallel and do not intersect. The shortest distance between them is the length of the unique line segment that is perpendicular to both lines.
• Vector Form of Lines: A line passing through a point with position vector $\vec{a}$ and having a direction vector $\vec{b}$ can be represented as $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter.
• Shortest Distance Formula: For two skew lines $L_1: \vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $L_2: \vec{r} = \vec{a_2} + \mu \vec{b_2}$, the shortest distance $d$ is given by: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ This formula essentially calculates the scalar projection of the vector connecting a point on $L_1$ to a point on $L_2$ onto the common normal vector to both lines.

2. Step-by-Step Solution

Step 1: Extract Position and Direction Vectors from the Line Equations

The lines are given in Cartesian form: $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$. From this, a point on the line is $(x_0, y_0, z_0)$ and its direction vector is $(a, b, c)$.

For Line 1 ($L_1$): ${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$

• Position vector of a point on $L_1$: $\vec{a_1} = 3\hat{i} + 8\hat{j} + 3\hat{k}$
• Direction vector of $L_1$: $\vec{b_1} = 3\hat{i} - \hat{j} + \hat{k}$

For Line 2 ($L_2$): ${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$

• Important Note: Remember to convert $x+k$ to $x-(-k)$ for the point's coordinate.
• Position vector of a point on $L_2$: $\vec{a_2} = -3\hat{i} - 7\hat{j} + 6\hat{k}$
• Direction vector of $L_2$: $\vec{b_2} = -3\hat{i} + 2\hat{j} + 4\hat{k}$

Step 2: Calculate the Vector Connecting the Points ($\vec{a_2} - \vec{a_1}$)

This vector connects a specific point on $L_1$ to a specific point on $L_2$. $\vec{a_2} - \vec{a_1} = (-3\hat{i} - 7\hat{j} + 6\hat{k}) - (3\hat{i} + 8\hat{j} + 3\hat{k})$ $\vec{a_2} - \vec{a_1} = (-3 - 3)\hat{i} + (-7 - 8)\hat{j} + (6 - 3)\hat{k}$ $\vec{a_2} - \vec{a_1} = -6\hat{i} - 15\hat{j} + 3\hat{k}$

Step 3: Calculate the Cross Product of the Direction Vectors ($\vec{b_1} \times \vec{b_2}$)

This cross product gives a vector that is perpendicular to both direction vectors, and therefore perpendicular to both lines. This vector represents the direction of the shortest distance. $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix}$ $= \hat{i}((-1)(4) - (1)(2)) - \hat{j}((3)(4) - (1)(-3)) + \hat{k}((3)(2) - (-1)(-3))$ $= \hat{i}(-4 - 2) - \hat{j}(12 - (-3)) + \hat{k}(6 - 3)$ $= \hat{i}(-6) - \hat{j}(15) + \hat{k}(3)$ $= -6\hat{i} - 15\hat{j} + 3\hat{k}$

• Observation: Notice that $\vec{a_2} - \vec{a_1}$ is identical to $\vec{b_1} \times \vec{b_2}$ in this particular problem. This is a special case.

Step 4: Calculate the Magnitude of the Cross Product ($|\vec{b_1} \times \vec{b_2}|$)

This magnitude will be the denominator in our formula. $|\vec{b_1} \times \vec{b_2}| = |-6\hat{i} - 15\hat{j} + 3\hat{k}|$ $= \sqrt{(-6)^2 + (-15)^2 + (3)^2}$ $= \sqrt{36 + 225 + 9}$ $= \sqrt{270}$ To simplify the radical: $270 = 9 \times 30$. $|\vec{b_1} \times \vec{b_2}| = \sqrt{9 \times 30} = 3\sqrt{30}$

Step 5: Calculate the Scalar Triple Product (Numerator of the Formula)

This is $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$. Since we observed that $\vec{a_2} - \vec{a_1} = \vec{b_1} \times \vec{b_2} = -6\hat{i} - 15\hat{j} + 3\hat{k}$, let's call this vector $\vec{V}$. Then the numerator is $\vec{V} \cdot \vec{V} = |\vec{V}|^2$. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-6\hat{i} - 15\hat{j} + 3\hat{k}) \cdot (-6\hat{i} - 15\hat{j} + 3\hat{k})$ $= (-6)(-6) + (-15)(-15) + (3)(3)$ $= 36 + 225 + 9$ $= 270$ Alternatively, we already found $|\vec{V}| = \sqrt{270}$, so $|\vec{V}|^2 = (\sqrt{270})^2 = 270$.

Step 6: Apply the Shortest Distance Formula

Substitute the calculated numerator and denominator into the formula: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ $d = \left| \frac{270}{3\sqrt{30}} \right|$ $d = \frac{270}{3\sqrt{30}}$ $d = \frac{90}{\sqrt{30}}$ To rationalize the denominator, multiply the numerator and denominator by $\sqrt{30}$: $d = \frac{90}{\sqrt{30}} \times \frac{\sqrt{30}}{\sqrt{30}}$ $d = \frac{90\sqrt{30}}{30}$ $d = 3\sqrt{30}$

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful when extracting coordinates from Cartesian form ($x+k$ means $x_0 = -k$) and during vector arithmetic (especially cross products and dot products).
• Parallel Lines Check: Always check if the direction vectors $\vec{b_1}$ and $\vec{b_2}$ are proportional. If they are, the lines are parallel (or coincident), and a different formula for shortest distance applies: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \times \vec{b}}{|\vec{b}|} \right|$. In this problem, $\vec{b_1}$ and $\vec{b_2}$ are not proportional, confirming they are skew.
• Intersecting Lines Check: If the scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ is zero, it means the lines are coplanar. If they are not parallel, they must intersect, and the shortest distance is zero. Here, the numerator is 270, so they do not intersect.

4. Summary

We systematically applied the formula for the shortest distance between two skew lines. First, we correctly identified the position and direction vectors for both lines. Then, we calculated the necessary components: the vector connecting points on the lines, the cross product of the direction vectors, its magnitude, and the scalar triple product. Substituting these values into the formula yielded the shortest distance.

The shortest distance between the given lines is $3\sqrt{30}$.

5. Final Answer

The final answer is $\boxed{3\sqrt{30}}$, which corresponds to option (C).