1. Key Concepts and Formulas

The problem revolves around finding the shortest distance between two skew lines. Skew lines are non-parallel and non-intersecting lines in 3D space.

• Vector Form of a Line: A line passing through a point with position vector $\vec{a}$ and parallel to a direction vector $\vec{d}$ can be represented as $\vec{r} = \vec{a} + t\vec{d}$, where $t$ is a scalar parameter. If a line is given in Cartesian form $\frac{x-x_0}{d_x} = \frac{y-y_0}{d_y} = \frac{z-z_0}{d_z}$, then a point on the line is $(x_0, y_0, z_0)$ (so $\vec{a} = x_0\hat{i} + y_0\hat{j} + z_0\hat{k}$) and its direction vector is $\vec{d} = d_x\hat{i} + d_y\hat{j} + d_z\hat{k}$.

• Shortest Distance Formula for Skew Lines: If two lines are given by $\vec{r_1} = \vec{a_1} + t\vec{d_1}$ and $\vec{r_2} = \vec{a_2} + s\vec{d_2}$, the shortest distance $D$ between them is given by: $D = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})}{|\vec{d_1} \times \vec{d_2}|} \right|$ This formula essentially calculates the scalar projection of the vector connecting any two points on the lines onto their common perpendicular direction.

• Vector Operations: The formula requires vector subtraction, cross product, dot product, and magnitude calculations.

2. Step-by-Step Solution

Step 1: Identify Position and Direction Vectors for Each Line

We are given the lines in Cartesian form. Our first step is to extract the position vector of a point on each line ($\vec{a_1}, \vec{a_2}$) and their respective direction vectors ($\vec{d_1}, \vec{d_2}$).

• Line 1 ($L_1$): $\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$

  • From the form $\frac{x-x_0}{d_x}$, we identify $x_0 = \lambda$, $y_0 = 3$, $z_0 = -6$.
  • So, a point on $L_1$ is $(\lambda, 3, -6)$, and its position vector is $\vec{a_1} = \lambda\hat{i} + 3\hat{j} - 6\hat{k}$.
  • The direction numbers are $d_x = 0$, $d_y = 4$, $d_z = 1$.
  • So, the direction vector of $L_1$ is $\vec{d_1} = 0\hat{i} + 4\hat{j} + 1\hat{k} = 4\hat{j} + \hat{k}$.

• Line 2 ($L_2$): $\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$

  • From the form $\frac{x-x_0}{d_x}$, we identify $x_0 = -\lambda$ (since $x+\lambda = x-(-\lambda)$), $y_0 = 0$, $z_0 = 6$.
  • So, a point on $L_2$ is $(-\lambda, 0, 6)$, and its position vector is $\vec{a_2} = -\lambda\hat{i} + 0\hat{j} + 6\hat{k} = -\lambda\hat{i} + 6\hat{k}$.
  • The direction numbers are $d_x = 3$, $d_y = -4$, $d_z = 0$.
  • So, the direction vector of $L_2$ is $\vec{d_2} = 3\hat{i} - 4\hat{j} + 0\hat{k} = 3\hat{i} - 4\hat{j}$.

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

This vector connects a point on $L_1$ to a point on $L_2$. It forms the basis for finding the shortest distance. $\vec{a_2} - \vec{a_1} = (-\lambda\hat{i} + 6\hat{k}) - (\lambda\hat{i} + 3\hat{j} - 6\hat{k})$ $= (-\lambda - \lambda)\hat{i} + (0 - 3)\hat{j} + (6 - (-6))\hat{k}$ $= -2\lambda\hat{i} - 3\hat{j} + 12\hat{k}$

Step 3: Calculate the Cross Product of Direction Vectors ($\vec{d_1} \times \vec{d_2}$) and its Magnitude

The cross product $\vec{d_1} \times \vec{d_2}$ gives a vector perpendicular to both direction vectors, which is the direction of the common perpendicular between the two lines. Its magnitude will be the denominator in the distance formula.

• Cross Product: $\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & 1 \\ 3 & -4 & 0 \end{vmatrix}$ $= \hat{i}(4 \cdot 0 - 1 \cdot (-4)) - \hat{j}(0 \cdot 0 - 1 \cdot 3) + \hat{k}(0 \cdot (-4) - 4 \cdot 3)$ $= \hat{i}(0 + 4) - \hat{j}(0 - 3) + \hat{k}(0 - 12)$ $= 4\hat{i} + 3\hat{j} - 12\hat{k}$

• Magnitude of Cross Product: $|\vec{d_1} \times \vec{d_2}| = \sqrt{4^2 + 3^2 + (-12)^2}$ $= \sqrt{16 + 9 + 144} = \sqrt{169} = 13$

Step 4: Calculate the Scalar Triple Product (Numerator of the Formula)

This is the dot product of $(\vec{a_2} - \vec{a_1})$ with $(\vec{d_1} \times \vec{d_2})$. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2}) = (-2\lambda\hat{i} - 3\hat{j} + 12\hat{k}) \cdot (4\hat{i} + 3\hat{j} - 12\hat{k})$ $= (-2\lambda)(4) + (-3)(3) + (12)(-12)$ $= -8\lambda - 9 - 144$ $= -8\lambda - 153$

Step 5: Apply the Shortest Distance Formula and Solve for $\lambda$

We are given that the shortest distance $D = 13$. Substitute the calculated values into the formula: $D = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})}{|\vec{d_1} \times \vec{d_2}|} \right|$ $13 = \left| \frac{-8\lambda - 153}{13} \right|$

Multiply both sides by 13 to remove the denominator: $13 \times 13 = |-8\lambda - 153|$ $169 = |-8\lambda - 153|$

To solve an absolute value equation $|X| = k$, we consider two cases: $X = k$ or $X = -k$.

• Case 1: $-8\lambda - 153 = 169$ $-8\lambda = 169 + 153$ $-8\lambda = 322$ $\lambda = -\frac{322}{8}$ $\lambda = -\frac{161}{4}$

• Case 2: $-8\lambda - 153 = -169$ $-8\lambda = -169 + 153$ $-8\lambda = -16$ $\lambda = \frac{-16}{-8}$ $\lambda = 2$

The set $S$ of all values of $\lambda$ is $S = \left\{ -\frac{161}{4}, 2 \right\}$.

Step 6: Calculate the Final Expression ($8\left|\sum\limits_{\lambda \in S} \lambda\right|$)

First, find the sum of all values of $\lambda$ in $S$: $\sum_{\lambda \in S} \lambda = -\frac{161}{4} + 2$ To add these, find a common denominator: $\sum_{\lambda \in S} \lambda = -\frac{161}{4} + \frac{8}{4} = -\frac{153}{4}$

Now, substitute this sum into the expression $8\left|\sum\limits_{\lambda \in S} \lambda\right|$: $8\left|\sum_{\lambda \in S} \lambda\right| = 8 \left| -\frac{153}{4} \right|$ Since $|-x| = x$: $= 8 \times \frac{153}{4}$ $= 2 \times 153$ $= 306$

3. Common Mistakes & Tips

• Sign Errors in Identifying Points: Be very careful when extracting $(x_0, y_0, z_0)$ from Cartesian forms like $x+\lambda$ (which means $x-(-\lambda)$, so the coordinate is $-\lambda$).
• Vector Calculation Errors: Cross products and dot products involve several multiplications and subtractions. Double-check these calculations, especially with negative signs.
• Absolute Value Handling: Remember the absolute value in the shortest distance formula. When solving $|X|=k$, always consider both $X=k$ and $X=-k$ to find all possible values of the variable.
• Understanding the Denominator: If $|\vec{d_1} \times \vec{d_2}|$ is zero, the lines are parallel (or coincident), and the shortest distance formula is different. In this problem, it was non-zero, indicating skew lines.

4. Summary

This problem required the application of the shortest distance formula between two skew lines. We first extracted the position vectors and direction vectors from the given Cartesian equations. Then, we systematically calculated the necessary vector quantities: the difference of position vectors, the cross product of direction vectors, and their dot product. Substituting these into the shortest distance formula and setting it equal to 13 led to an absolute value equation. Solving this equation yielded two possible values for $\lambda$. Finally, we summed these values, took the absolute value, and multiplied by 8 to arrive at the required answer.

5. Final Answer

The final value of $8\left|\sum\limits_{\lambda \in S} \lambda\right|$ is 306. The final answer is $\boxed{\text{306}}$, which corresponds to option (A).