Key Concepts and Formulas

1. Vector Equation of a Line: A line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ can be represented as $\vec{r} = \vec{a} + t\vec{b}$, where $t$ is a scalar parameter.
2. Shortest Distance Between Two Skew Lines: For two lines $L_1: \vec{r} = \vec{a_1} + t\vec{b_1}$ and $L_2: \vec{r} = \vec{a_2} + s\vec{b_2}$, the shortest distance $d$ between them is given by: $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$
3. Distance of a Point from a Line: To find the distance from a point $P(\vec{p})$ to a line $L: \vec{r} = \vec{a} + t\vec{b}$, we find the foot of the perpendicular $Q$ on $L$. The vector $\vec{PQ}$ must be perpendicular to the line's direction vector $\vec{b}$, i.e., $\vec{PQ} \cdot \vec{b} = 0$. Once $Q$ is found, the distance is $PQ = |\vec{PQ}|$.

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Step-by-Step Solution

Step 1: Express Lines $L_1$ and $L_2$ in Vector Form

We begin by converting the given information about the lines into their standard vector equations. This is essential for applying the shortest distance formula.

• For Line $L_1$:

  • It passes through the point $(1,2,3)$, so its position vector $\vec{a_1}$ is $\hat{i} + 2\hat{j} + 3\hat{k}$.
  • It is parallel to the $z$-axis, so its direction vector $\vec{b_1}$ is $\hat{k}$.
  • Thus, the vector equation of $L_1$ is: $\vec{r}_1 = (\hat{i} + 2\hat{j} + 3\hat{k}) + t\hat{k}$

• For Line $L_2$:

  • It passes through the point $(\lambda, 5,6)$, so its position vector $\vec{a_2}$ is $\lambda\hat{i} + 5\hat{j} + 6\hat{k}$.
  • It is parallel to the $y$-axis, so its direction vector $\vec{b_2}$ is $\hat{j}$.
  • Thus, the vector equation of $L_2$ is: $\vec{r}_2 = (\lambda\hat{i} + 5\hat{j} + 6\hat{k}) + s\hat{j}$

Step 2: Calculate the Shortest Distance between $L_1$ and $L_2$ and Determine $\lambda$

We are given that the shortest distance between $L_1$ and $L_2$ is 3. We will use the formula for the shortest distance between two skew lines.

• Calculate $\vec{a_2} - \vec{a_1}$: This vector connects a point on $L_1$ to a point on $L_2$. $\vec{a_2} - \vec{a_1} = (\lambda\hat{i} + 5\hat{j} + 6\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = (\lambda-1)\hat{i} + 3\hat{j} + 3\hat{k}$

• Calculate $\vec{b_1} \times \vec{b_2}$: This vector gives the direction of the common perpendicular between the two lines. $\vec{b_1} \times \vec{b_2} = \hat{k} \times \hat{j} = -\hat{i}$

• Calculate $|\vec{b_1} \times \vec{b_2}|$: This is the magnitude of the common perpendicular direction vector. $|\vec{b_1} \times \vec{b_2}| = |-\hat{i}| = 1$

• Calculate $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$: This scalar triple product represents the projection of the connecting vector onto the common perpendicular. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = ((\lambda-1)\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-\hat{i}) = -(\lambda-1) = 1-\lambda$

• Apply the shortest distance formula: $d = \frac{|1-\lambda|}{1} = |1-\lambda|$ Given $d=3$, we have: $|1-\lambda| = 3$ This yields two possible values for $\lambda$: $1-\lambda = 3 \quad \text{or} \quad 1-\lambda = -3$ $\lambda = -2 \quad \text{or} \quad \lambda = 4$

Step 3: Identify $\lambda_1$ and $\lambda_2$

The problem states that for $\lambda=\lambda_1, \lambda_2$, the shortest distance is 3, and that $\lambda_2 < \lambda_1$. From our calculated values, $-2$ and $4$:

• $\lambda_1 = 4$
• $\lambda_2 = -2$ This satisfies the condition $\lambda_2 < \lambda_1$.

Step 4: Identify the Target Point and Line for the Final Distance Calculation

We need to find the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from the line $L_1$.

• Target Point $P$: Substitute the values of $\lambda_1$ and $\lambda_2$: $P(4, -2, 7)$
• Target Line $L_1$: From Step 1, $L_1$ passes through $A(1,2,3)$ and has a direction vector $\vec{b_1} = \hat{k}$. A general point $Q$ on line $L_1$ can be represented as $(1, 2, 3+t)$.

Step 5: Calculate the Square of the Distance from Point $P(4, -2, 7)$ to Line $L_1$

To find the distance from point $P$ to line $L_1$, we find the foot of the perpendicular $Q$ from $P$ onto $L_1$.

• Form the vector $\vec{PQ}$: This vector connects the point $P$ to an arbitrary point $Q$ on $L_1$. $\vec{PQ} = (1-4)\hat{i} + (2-(-2))\hat{j} + (3+t-7)\hat{k}$ $\vec{PQ} = -3\hat{i} + 4\hat{j} + (t-4)\hat{k}$

• Apply the perpendicularity condition: The vector $\vec{PQ}$ must be perpendicular to the direction vector of $L_1$, which is $\vec{b_1} = \hat{k}$. Their dot product must be zero. $\vec{PQ} \cdot \vec{b_1} = 0$ $(-3\hat{i} + 4\hat{j} + (t-4)\hat{k}) \cdot \hat{k} = 0$ $(t-4)(1) = 0 \implies t=4$

• Determine the vector $\vec{PQ}$ at the foot of the perpendicular: Substitute $t=4$ back into the expression for $\vec{PQ}$. $\vec{PQ} = -3\hat{i} + 4\hat{j} + (4-4)\hat{k} = -3\hat{i} + 4\hat{j}$

• Calculate the distance $PQ$: The distance is the magnitude of this vector. $PQ = |\vec{PQ}| = \sqrt{(-3)^2 + 4^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

• Calculate the square of the distance: The question asks for the square of the distance. $(PQ)^2 = 5^2 = 25$

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Common Mistakes & Tips

• Vector Algebra Precision: Be extremely careful with signs and components during cross products and dot products, as small errors can propagate.
• Absolute Value: Always remember the absolute value in the shortest distance formula, $|1-\lambda|=3$ leads to two distinct values for $\lambda$.
• Condition on $\lambda_1, \lambda_2$: Pay close attention to conditions like $\lambda_2 < \lambda_1$ to correctly assign the values.
• Final Requirement: Double-check whether the question asks for the distance or the square of the distance.

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Summary

This problem effectively tests the understanding of 3D geometry concepts. We first defined the two lines in vector form using the given points and parallel axes. Then, we applied the shortest distance formula between skew lines, which involved calculating a vector difference, a cross product, and a dot product. This led to an equation involving $\lambda$, yielding two solutions. By using the condition $\lambda_2 < \lambda_1$, we correctly identified $\lambda_1=4$ and $\lambda_2=-2$. Finally, we calculated the distance from the derived point $(\lambda_1, \lambda_2, 7)$ to line $L_1$ by finding the foot of the perpendicular, and then squared this distance as requested.

The final answer is $\boxed{\text{25}}$, which corresponds to option (A).