1. Key Concepts and Formulas

• Shortest Distance Between Skew Lines: The shortest distance $D$ between two skew lines (lines that are neither parallel nor intersecting), represented in vector form as $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$, is given by the formula: $D = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ Here:
  • $\vec{a_1}$ and $\vec{a_2}$ are the position vectors of arbitrary points on the first and second lines, respectively.
  • $\vec{b_1}$ and $\vec{b_2}$ are the direction vectors of the first and second lines, respectively.
  • The term $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ is the scalar triple product (STP).
  • The term $|\vec{b_1} \times \vec{b_2}|$ is the magnitude of the cross product of the direction vectors.
• Vector Form of a Line: A line in Cartesian form $\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}$ can be written in vector form as $\vec{r} = (x_0\hat{i} + y_0\hat{j} + z_0\hat{k}) + t(l\hat{i} + m\hat{j} + n\hat{k})$.
• Scalar Triple Product (STP) as a Determinant: The scalar triple product $(\vec{u} \cdot (\vec{v} \times \vec{w}))$ can be computed using a determinant: $(\vec{u} \cdot (\vec{v} \times \vec{w})) = \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix}$

2. Step-by-Step Solution

Step 1: Extract Position and Direction Vectors from Given Lines We convert the given Cartesian equations of the lines into their vector forms $\vec{r} = \vec{a} + t\vec{b}$.

For Line 1 ($L_1$): $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$

• The position vector $\vec{a_1}$ (representing a point $(x_0, y_0, z_0)$ on $L_1$) is obtained from $(4, -1, 0)$. So, $\vec{a_1} = 4\hat{i} - \hat{j} + 0\hat{k}$.
• The direction vector $\vec{b_1}$ (representing the direction ratios $(l, m, n)$ of $L_1$) is obtained from $(1, 2, -3)$. So, $\vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k}$.

For Line 2 ($L_2$): $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$

• The position vector $\vec{a_2}$ (representing a point $(x_0, y_0, z_0)$ on $L_2$) is obtained from $(\lambda, -1, 2)$. So, $\vec{a_2} = \lambda\hat{i} - \hat{j} + 2\hat{k}$.
• The direction vector $\vec{b_2}$ (representing the direction ratios $(l, m, n)$ of $L_2$) is obtained from $(2, 4, -5)$. So, $\vec{b_2} = 2\hat{i} + 4\hat{j} - 5\hat{k}$.

Important Check for Parallelism: Before proceeding with the skew lines formula, we must check if the lines are parallel. Lines are parallel if their direction vectors are proportional. Here, $\vec{b_1} = \langle 1, 2, -3 \rangle$ and $\vec{b_2} = \langle 2, 4, -5 \rangle$. Since $\frac{1}{2} = \frac{2}{4} \neq \frac{-3}{-5}$, the direction vectors are not proportional. This confirms that the lines are not parallel, and thus they are skew lines. The chosen formula is appropriate.

Step 2: Calculate the Vector Connecting Points on the Lines, $\vec{a_2} - \vec{a_1}$ This vector connects the specific point $A_1(4, -1, 0)$ on $L_1$ to $A_2(\lambda, -1, 2)$ on $L_2$. $\vec{a_2} - \vec{a_1} = (\lambda\hat{i} - \hat{j} + 2\hat{k}) - (4\hat{i} - \hat{j} + 0\hat{k})$ $\vec{a_2} - \vec{a_1} = (\lambda - 4)\hat{i} + (-1 - (-1))\hat{j} + (2 - 0)\hat{k}$ $\vec{a_2} - \vec{a_1} = (\lambda - 4)\hat{i} + 0\hat{j} + 2\hat{k}$

Step 3: Calculate the Cross Product of Direction Vectors, $\vec{b_1} \times \vec{b_2}$ This vector is perpendicular to both lines, indicating the direction of the shortest distance. $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix}$ Expanding the determinant: $= \hat{i}((2)(-5) - (-3)(4)) - \hat{j}((1)(-5) - (-3)(2)) + \hat{k}((1)(4) - (2)(2))$ $= \hat{i}(-10 + 12) - \hat{j}(-5 + 6) + \hat{k}(4 - 4)$ $= 2\hat{i} - 1\hat{j} + 0\hat{k}$

Step 4: Calculate the Magnitude of the Cross Product, $|\vec{b_1} \times \vec{b_2}|$ This magnitude forms the denominator of the shortest distance formula. $|\vec{b_1} \times \vec{b_2}| = |2\hat{i} - \hat{j} + 0\hat{k}| = \sqrt{(2)^2 + (-1)^2 + (0)^2}$ $= \sqrt{4 + 1 + 0} = \sqrt{5}$

Step 5: Calculate the Scalar Triple Product (STP), $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ This STP forms the numerator of the shortest distance formula. We can compute it by taking the dot product of the results from Step 2 and Step 3. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = ((\lambda - 4)\hat{i} + 0\hat{j} + 2\hat{k}) \cdot (2\hat{i} - \hat{j} + 0\hat{k})$ $= (\lambda - 4)(2) + (0)(-1) + (2)(0)$ $= 2\lambda - 8 + 0 + 0$ $= 2\lambda - 8$ Alternatively, using the determinant form for the scalar triple product: $\begin{vmatrix} \lambda-4 & 0 & 2 \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = (\lambda-4) \begin{vmatrix} 2 & -3 \\ 4 & -5 \end{vmatrix} - 0 \begin{vmatrix} 1 & -3 \\ 2 & -5 \end{vmatrix} + 2 \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix}$ $= (\lambda-4)((2)(-5) - (-3)(4)) - 0 + 2((1)(4) - (2)(2))$ $= (\lambda-4)(-10 + 12) + 2(4 - 4)$ $= (\lambda-4)(2) + 2(0) = 2\lambda - 8$ Both methods confirm the numerator is $2\lambda - 8$.

Step 6: Apply the Shortest Distance Formula and Solve for $\lambda$ Now, we substitute the calculated components into the shortest distance formula: $D = \left| \frac{2\lambda - 8}{\sqrt{5}} \right|$ The problem states that the shortest distance $D = \frac{6}{\sqrt{5}}$. So, we set up the equation: $\left| \frac{2\lambda - 8}{\sqrt{5}} \right| = \frac{6}{\sqrt{5}}$ Since the denominators are equal and positive, we can equate the numerators (with the absolute value): $|2\lambda - 8| = 6$ Factor out 2 from the absolute value expression: $|2(\lambda - 4)| = 6$ $2|\lambda - 4| = 6$ Divide by 2: $|\lambda - 4| = 3$ This absolute value equation leads to two possible cases: Case 1: $\lambda - 4 = 3$ $\lambda = 3 + 4 \implies \lambda = 7$ Case 2: $\lambda - 4 = -3$ $\lambda = -3 + 4 \implies \lambda = 1$ The possible values of $\lambda$ are $7$ and $1$.

Step 7: Find the Sum of All Possible Values of $\lambda$ The problem asks for the sum of all possible values of $\lambda$. Sum $= 7 + 1 = 8$.

3. Common Mistakes & Tips

• Sign Errors: A common pitfall is making sign errors when extracting point coordinates (e.g., $y+1$ implies $y_0 = -1$) or during vector operations like cross products and dot products. Double-check all signs.
• Forgetting Absolute Value: The shortest distance must always be non-negative. Forgetting the absolute value in the formula, i.e., $D = \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|}$, could lead to missing one of the possible values for the unknown variable (like $\lambda$ here).
• Arithmetic Errors: Calculations involving determinants, products, and sums can be prone to arithmetic mistakes. It's wise to perform these calculations carefully and re-check them.

4. Summary

To find the sum of all possible values of $\lambda$, we utilized the formula for the shortest distance between two skew lines. We first correctly identified the position and direction vectors from the given Cartesian equations of the lines. After verifying that the lines were indeed skew (not parallel), we systematically calculated the necessary components: $\vec{a_2} - \vec{a_1}$, $\vec{b_1} \times \vec{b_2}$, and their magnitudes and dot products. Substituting these into the shortest distance formula and equating it to the given distance $\frac{6}{\sqrt{5}}$, we obtained an absolute value equation: $|\lambda - 4| = 3$. Solving this equation yielded two possible values for $\lambda$, which are $7$ and $1$. The sum of these values is $8$.

5. Final Answer The final answer is $\boxed{8}$, which corresponds to option (D).