Key Concepts and Formulas

• Direction of the Line of Shortest Distance (LSD) between two skew lines: If two lines are given by $\vec{r_1} = \vec{a_1} + \lambda \vec{d_1}$ and $\vec{r_2} = \vec{a_2} + \mu \vec{d_2}$, the line of shortest distance is perpendicular to both $\vec{d_1}$ and $\vec{d_2}$. Therefore, its direction vector is given by $\vec{N} = \vec{d_1} \times \vec{d_2}$.
• Equation of a Line: A line passing through a point $(x_0, y_0, z_0)$ with a direction vector $(a, b, c)$ can be expressed parametrically as $x = x_0 + ta$, $y = y_0 + tb$, $z = z_0 + tc$, where $t$ is a scalar parameter.
• Intersection of a Line and a Plane: To find the point where a line intersects a plane, substitute the parametric coordinates of a general point on the line into the Cartesian equation of the plane and solve for the parameter $t$.
• Distance between Two Points: The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3D space is given by the formula: $\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Step-by-Step Solution

This problem asks for the distance of a point from a plane, measured along a specific direction rather than perpendicularly. This direction is defined by the line of shortest distance (LSD) between two other given lines. Our strategy involves four main steps: finding the direction, constructing a line through the given point in that direction, finding where this line intersects the plane, and finally calculating the distance between the two points.

Step 1: Determine the Direction Vector of the Line of Shortest Distance (LSD) Why: The problem statement explicitly requires the distance to be measured "parallel to the line of the shortest distance". Therefore, identifying this direction vector is the fundamental first step, as it defines the orientation of the path from our starting point to the plane.

The two given lines are:

• $L_1: \vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})$
• $L_2: \vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})$

From these equations, we extract their respective direction vectors:

• Direction vector of $L_1$: $\vec{d_1} = 2\hat{i} + 0\hat{j} + 1\hat{k}$
• Direction vector of $L_2$: $\vec{d_2} = 1\hat{i} - 1\hat{j} + 1\hat{k}$

The direction vector of the line of shortest distance, which we'll denote as $\vec{D}$, is perpendicular to both $\vec{d_1}$ and $\vec{d_2}$. We calculate this using the cross product: $\vec{D} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & -1 & 1 \\ \end{vmatrix}$ Expanding the determinant: $\vec{D} = \hat{i}((0)(1) - (1)(-1)) - \hat{j}((2)(1) - (1)(1)) + \hat{k}((2)(-1) - (0)(1))$ $\vec{D} = \hat{i}(0 + 1) - \hat{j}(2 - 1) + \hat{k}(-2 - 0)$ $\vec{D} = \hat{i} - \hat{j} - 2\hat{k}$ So, the direction vector we need is $(1, -1, -2)$.

Step 2: Form the Line Passing Through Point A Parallel to $\vec{D}$ Why: This step creates the specific "path" along which we are required to measure the distance. This line starts at the given point $A$ and extends in the direction $\vec{D}$ until it reaches the plane.

The given point is $A(-1, 2, 3)$. The direction vector we found is $\vec{D} = (1, -1, -2)$.

The parametric equation of the line $L_A$ passing through $A$ and parallel to $\vec{D}$ is: $\frac{x - (-1)}{1} = \frac{y - 2}{-1} = \frac{z - 3}{-2} = t$ From this parametric form, any general point $P$ on this line can be represented by its coordinates in terms of the parameter $t$: $P(t - 1, -t + 2, -2t + 3)$

Step 3: Find the Intersection Point P of Line $L_A$ and the Plane Why: This intersection point, $P$, represents the exact location on the plane where the path from point $A$ (along direction $\vec{D}$) terminates. The distance we seek is the length of this path from $A$ to $P$.

First, convert the given plane equation from vector form to its Cartesian equivalent for easier substitution. The plane is given by $\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=10$. Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Then, $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i}-2 \hat{j}+3 \hat{k}) = 10$, which simplifies to: $x - 2y + 3z = 10$ Now, substitute the parametric coordinates of the general point $P(t - 1, -t + 2, -2t + 3)$ from Step 2 into the Cartesian equation of the plane: $(t - 1) - 2(-t + 2) + 3(-2t + 3) = 10$ Solve this equation for $t$: $t - 1 + 2t - 4 - 6t + 9 = 10$ Combine the $t$ terms and the constant terms: $(t + 2t - 6t) + (-1 - 4 + 9) = 10$ $-3t + 4 = 10$ $-3t = 6$ $t = -2$ Now, substitute the value of $t = -2$ back into the parametric coordinates of $P$ to find its exact coordinates:

• $x_P = (-2) - 1 = -3$
• $y_P = -(-2) + 2 = 2 + 2 = 4$
• $z_P = -2(-2) + 3 = 4 + 3 = 7$ So, the intersection point is $P(-3, 4, 7)$.

Step 4: Calculate the Distance between Point A and Point P Why: This is the final calculation that directly provides the answer to the problem. The distance between the initial point $A$ and the intersection point $P$ is precisely the distance of point $A$ from the plane, measured along the specified direction.

We have our starting point $A(-1, 2, 3)$ and the intersection point $P(-3, 4, 7)$. Using the 3D distance formula: $\text{Distance } AP = \sqrt{(x_P - x_A)^2 + (y_P - y_A)^2 + (z_P - z_A)^2}$ $AP = \sqrt{(-3 - (-1))^2 + (4 - 2)^2 + (7 - 3)^2}$ $AP = \sqrt{(-3 + 1)^2 + (2)^2 + (4)^2}$ $AP = \sqrt{(-2)^2 + (2)^2 + (4)^2}$ $AP = \sqrt{4 + 4 + 16}$ $AP = \sqrt{24}$ To simplify $\sqrt{24}$, we factor out the perfect square $4$: $AP = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2\sqrt{6}$

Common Mistakes & Tips

• Misinterpreting "Distance": Always carefully read if the problem asks for perpendicular distance (which has a direct formula) or distance parallel to a specific direction (which requires the line-plane intersection method used here).
• Cross Product Accuracy: Errors in calculating the cross product can propagate through the entire solution. Double-check your determinant expansion and the signs of the components.
• Algebraic Vigilance: Solving for the parameter $t$ and subsequently finding the intersection point's coordinates requires careful algebraic manipulation, especially with negative numbers.
• Conversion between Forms: Be proficient in converting between vector and Cartesian forms for both lines and planes, as this is often necessary for substitution.

Summary To determine the distance of a point from a plane parallel to a given direction, the process involves four main steps: first, find the direction vector of the specified line (in this case, the line of shortest distance); second, construct a line passing through the given point and parallel to this direction; third, calculate the point of intersection of this constructed line with the plane; and finally, compute the distance between the initial point and this intersection point. By following these steps meticulously, the distance is found to be $2\sqrt{6}$.

Final Answer The final answer is $\boxed{2 \sqrt{6}}$, which corresponds to option (C).