Key Concepts and Formulas

• Line in Vector Form: A line passing through a point with position vector $\vec{a}$ and having a direction vector $\vec{b}$ can be represented as $\vec{r} = \vec{a} + t\vec{b}$, where $t$ is a scalar parameter. From its Cartesian form $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$, we have $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and $\vec{b} = l\hat{i} + m\hat{j} + n\hat{k}$.
• Shortest Distance Between Two Skew Lines: For two skew lines given by $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$, the shortest distance $d$ is given by the formula: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$
• Vieta's Formulas for Quadratic Equations: For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$.

Step-by-Step Solution

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

We are given two lines in Cartesian form. To use the shortest distance formula, we first convert them into their vector forms $\vec{r} = \vec{a} + t\vec{b}$.

For the first line: ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over \lambda }$

• This line passes through the point $(1, 2, 3)$. So, its position vector is $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$.
• The direction ratios are $(2, 3, \lambda)$. So, its direction vector is $\vec{b_1} = 2\hat{i} + 3\hat{j} + \lambda\hat{k}$.

For the second line: ${{x - 2} \over 1} = {{y - 4} \over 4} = {{z - 5} \over 5}$

• This line passes through the point $(2, 4, 5)$. So, its position vector is $\vec{a_2} = 2\hat{i} + 4\hat{j} + 5\hat{k}$.
• The direction ratios are $(1, 4, 5)$. So, its direction vector is $\vec{b_2} = \hat{i} + 4\hat{j} + 5\hat{k}$.

Step 2: Calculating Necessary Vector Components for the Shortest Distance Formula

We need to calculate $\vec{a_2} - \vec{a_1}$, $\vec{b_1} \times \vec{b_2}$, and then their dot product and the magnitude of the cross product.

1. Calculate the vector connecting points on the two lines, $\vec{a_2} - \vec{a_1}$: This vector represents the displacement from a point on the first line to a point on the second line. $\vec{a_2} - \vec{a_1} = (2\hat{i} + 4\hat{j} + 5\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k})$ $\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$

2. Calculate the cross product of the direction vectors, $\vec{b_1} \times \vec{b_2}$: This vector is perpendicular to both direction vectors, and its direction gives the normal to the plane containing the shortest distance. $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & \lambda \\ 1 & 4 & 5 \end{vmatrix}$ $= \hat{i}(3 \times 5 - \lambda \times 4) - \hat{j}(2 \times 5 - \lambda \times 1) + \hat{k}(2 \times 4 - 3 \times 1)$ $= \hat{i}(15 - 4\lambda) - \hat{j}(10 - \lambda) + \hat{k}(8 - 3)$ $= (15 - 4\lambda)\hat{i} + (\lambda - 10)\hat{j} + 5\hat{k}$

3. Calculate the scalar triple product (numerator of the formula): $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ This dot product gives the projection of the connecting vector onto the normal vector, which is the shortest distance times the magnitude of the normal vector. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (\hat{i} + 2\hat{j} + 2\hat{k}) \cdot ((15 - 4\lambda)\hat{i} + (\lambda - 10)\hat{j} + 5\hat{k})$ $= 1(15 - 4\lambda) + 2(\lambda - 10) + 2(5)$ $= 15 - 4\lambda + 2\lambda - 20 + 10$ $= -2\lambda + 5$

4. Calculate the magnitude of the cross product (denominator of the formula): $|\vec{b_1} \times \vec{b_2}|$ This magnitude normalizes the scalar triple product to give the actual shortest distance. $|\vec{b_1} \times \vec{b_2}| = \sqrt{(15 - 4\lambda)^2 + (\lambda - 10)^2 + 5^2}$ $= \sqrt{(225 - 120\lambda + 16\lambda^2) + (\lambda^2 - 20\lambda + 100) + 25}$ $= \sqrt{17\lambda^2 - 140\lambda + 350}$

Step 3: Applying the Shortest Distance Formula

We are given that the shortest distance $d = \frac{1}{\sqrt{3}}$. Substituting the calculated components into the formula: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ $\frac{1}{\sqrt{3}} = \left| \frac{-2\lambda + 5}{\sqrt{17\lambda^2 - 140\lambda + 350}} \right|$

Step 4: Solving the Equation for $\lambda$

To solve for $\lambda$, we square both sides of the equation to eliminate the absolute value and the square roots: $\left( \frac{1}{\sqrt{3}} \right)^2 = \left( \frac{-2\lambda + 5}{\sqrt{17\lambda^2 - 140\lambda + 350}} \right)^2$ $\frac{1}{3} = \frac{(-2\lambda + 5)^2}{17\lambda^2 - 140\lambda + 350}$ Expand the numerator: $(-2\lambda + 5)^2 = (2\lambda - 5)^2 = 4\lambda^2 - 20\lambda + 25$. $\frac{1}{3} = \frac{4\lambda^2 - 20\lambda + 25}{17\lambda^2 - 140\lambda + 350}$ Cross-multiply to form a quadratic equation: $1(17\lambda^2 - 140\lambda + 350) = 3(4\lambda^2 - 20\lambda + 25)$ $17\lambda^2 - 140\lambda + 350 = 12\lambda^2 - 60\lambda + 75$ Rearrange the terms to form a standard quadratic equation $A\lambda^2 + B\lambda + C = 0$: $(17\lambda^2 - 12\lambda^2) + (-140\lambda + 60\lambda) + (350 - 75) = 0$ $5\lambda^2 - 80\lambda + 275 = 0$ Divide the entire equation by 5 to simplify: $\lambda^2 - 16\lambda + 55 = 0$

Step 5: Finding the Sum of Possible $\lambda$ Values

We have a quadratic equation $\lambda^2 - 16\lambda + 55 = 0$. Let the possible values of $\lambda$ be $\lambda_1$ and $\lambda_2$. According to Vieta's formulas, for a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$. In this equation, $a=1$, $b=-16$, and $c=55$. The sum of the possible values of $\lambda$ is $\lambda_1 + \lambda_2 = -(-16)/1 = 16$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful when extracting points and direction vectors from Cartesian form, and during vector operations (especially cross product and dot product). A single sign error can lead to an incorrect quadratic equation.
• Absolute Value: Remember the absolute value in the shortest distance formula. Squaring both sides is the correct approach to handle it, ensuring all terms are squared properly.
• Algebraic Simplification: Meticulously perform the algebraic expansions and simplifications, particularly when dealing with $\lambda$ inside square roots and squared terms.
• Vieta's Formulas: For problems asking for the sum or product of roots of a quadratic equation, Vieta's formulas offer a direct and efficient method, often saving time compared to finding individual roots.

Summary

This problem required us to find the parameter $\lambda$ using the shortest distance formula between two skew lines. We first converted the given Cartesian equations into vector form to identify the position and direction vectors. Then, we systematically calculated the necessary vector quantities: the difference of position vectors, the cross product of direction vectors, their dot product (for the numerator), and the magnitude of the cross product (for the denominator). Substituting these into the shortest distance formula and squaring both sides led to a quadratic equation in $\lambda$. Finally, using Vieta's formulas, we found the sum of all possible values of $\lambda$.

The final answer is $\boxed{\text{16}}$, which corresponds to option (A).