1. Key Concepts and Formulas

• Condition for a Line to Lie in a Plane: A line given by $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$ (with a point $P_1(x_1, y_1, z_1)$ and direction vector $\vec{b} = (l, m, n)$) lies in a plane $Ax+By+Cz=D$ (with normal vector $\vec{n} = (A, B, C)$) if and only if:

  1. The point $P_1$ on the line satisfies the plane equation, i.e., $Ax_1+By_1+Cz_1=D$.
  2. The direction vector of the line $\vec{b}$ is perpendicular to the normal vector of the plane $\vec{n}$, i.e., $\vec{b} \cdot \vec{n} = 0$, or $Al+Bm+Cn=0$.

• Shortest Distance Between Two Lines: For two lines $L_1: \vec{r} = \vec{a_1} + t\vec{b_1}$ and $L_2: \vec{r} = \vec{a_2} + s\vec{b_2}$:

  • If the lines are skew (non-parallel and non-intersecting), the shortest distance $d$ is given by: $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{||\vec{b_1} \times \vec{b_2}||}$
  • If the lines are coplanar (either parallel or intersecting), the shortest distance is 0 if they intersect, or calculated differently if they are parallel. If the scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ is zero, the lines are coplanar. If they are also non-parallel, they must intersect, and the shortest distance is 0.

2. Step-by-Step Solution

Part 1: Finding the value of $\lambda$

We are given the line $L_1: \frac{x - 3}{1} = \frac{y + 2}{-1} = \frac{z + \lambda}{-2}$ and the plane $P: 2x - 4y + 3z = 2$.

• Step 1.1: Identify a point and direction vector for line $L_1$. From the equation of $L_1$, a point on the line is $P_1(3, -2, -\lambda)$. The direction vector of $L_1$ is $\vec{b_1} = (1, -1, -2)$.

• Step 1.2: Identify the normal vector for the plane $P$. From the equation of plane $P$, the normal vector is $\vec{n} = (2, -4, 3)$.

• Step 1.3: Apply the first condition for $L_1$ to lie in the plane $P$. For $L_1$ to lie in $P$, the point $P_1(3, -2, -\lambda)$ must satisfy the plane's equation $2x - 4y + 3z = 2$. Substitute the coordinates of $P_1$: $2(3) - 4(-2) + 3(-\lambda) = 2$ $6 + 8 - 3\lambda = 2$ $14 - 3\lambda = 2$ $3\lambda = 12$ $\lambda = 4$

• Step 1.4: Apply the second condition for $L_1$ to lie in the plane $P$ (and verify). The direction vector of the line $\vec{b_1}$ must be perpendicular to the normal vector of the plane $\vec{n}$. Their dot product must be zero. Calculate $\vec{b_1} \cdot \vec{n}$: $(1, -1, -2) \cdot (2, -4, 3) = (1)(2) + (-1)(-4) + (-2)(3)$ $= 2 + 4 - 6 = 0$ Since the dot product is 0, this condition is satisfied. This confirms that for $\lambda=4$, the line $L_1$ lies in the plane $P$.

Part 2: Finding the shortest distance between the two lines

Now that we have $\lambda = 4$, the first line is $L_1: \frac{x - 3}{1} = \frac{y + 2}{-1} = \frac{z + 4}{-2}$. The second line is $L_2: \frac{x - 1}{12} = \frac{y}{9} = \frac{z}{4}$.

• Step 2.1: Extract points and direction vectors for both lines. For $L_1$: $\vec{a_1} = (3, -2, -4)$ and $\vec{b_1} = (1, -1, -2)$. For $L_2$: $\vec{a_2} = (1, 0, 0)$ and $\vec{b_2} = (12, 9, 4)$.

• Step 2.2: Check if line $L_2$ also lies in the same plane $P$. This step is crucial for understanding the nature of the lines. Point $P_2(1,0,0)$ on $L_2$. Substitute into plane $P: 2x - 4y + 3z = 2$: $2(1) - 4(0) + 3(0) = 2 - 0 + 0 = 2$ The point $P_2$ satisfies the plane equation. Now, check if $\vec{b_2}$ is perpendicular to $\vec{n}$: $\vec{b_2} \cdot \vec{n} = (12, 9, 4) \cdot (2, -4, 3) = (12)(2) + (9)(-4) + (4)(3)$ $= 24 - 36 + 12 = 0$ Since both conditions are met, the line $L_2$ also lies in the plane $2x - 4y + 3z = 2$.

• Step 2.3: Determine the relationship between $L_1$ and $L_2$. Both lines $L_1$ and $L_2$ lie in the same plane. Their direction vectors are $\vec{b_1} = (1, -1, -2)$ and $\vec{b_2} = (12, 9, 4)$. Since $\frac{1}{12} \ne \frac{-1}{9} \ne \frac{-2}{4}$, the direction vectors are not proportional, meaning the lines are not parallel. As both lines are coplanar and not parallel, they must intersect.

• Step 2.4: Conclude the shortest distance. The shortest distance between two intersecting lines is always 0.

  (For completeness, we can also use the shortest distance formula for skew lines and observe the result):

  • Calculate $(\vec{a_2} - \vec{a_1})$: $\vec{a_2} - \vec{a_1} = (1 - 3, 0 - (-2), 0 - (-4)) = (-2, 2, 4)$
  • Calculate the cross product $(\vec{b_1} \times \vec{b_2})$: $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & -2 \\ 12 & 9 & 4 \end{vmatrix} = \mathbf{i}((-1)(4) - (9)(-2)) - \mathbf{j}((1)(4) - (12)(-2)) + \mathbf{k}((1)(9) - (12)(-1))$ $= \mathbf{i}(-4 + 18) - \mathbf{j}(4 + 24) + \mathbf{k}(9 + 12) = 14\mathbf{i} - 28\mathbf{j} + 21\mathbf{k} = (14, -28, 21)$
  • Calculate the scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$: $(-2, 2, 4) \cdot (14, -28, 21) = (-2)(14) + (2)(-28) + (4)(21)$ $= -28 - 56 + 84 = -84 + 84 = 0$
  • Calculate the magnitude of the cross product $||\vec{b_1} \times \vec{b_2}||$: $||\vec{b_1} \times \vec{b_2}|| = ||(14, -28, 21)|| = \sqrt{14^2 + (-28)^2 + 21^2} = \sqrt{196 + 784 + 441} = \sqrt{1421}$
  • Substitute values into the shortest distance formula: $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{||\vec{b_1} \times \vec{b_2}||} = \frac{|0|}{\sqrt{1421}} = 0$

3. Common Mistakes & Tips

• Verifying both conditions for a line in a plane: Always check both that a point on the line satisfies the plane equation and that the line's direction vector is perpendicular to the plane's normal vector.
• Vector Arithmetic: Be meticulous with vector calculations, especially cross products and dot products, as sign errors are very common.
• Interpreting Shortest Distance: If the scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ is zero for two non-parallel lines, it implies that the lines are coplanar and thus intersect. The shortest distance between intersecting lines is always 0.

4. Summary

We first determined the value of $\lambda$ by applying the conditions for a line to lie within a plane. This yielded $\lambda=4$. Then, we analyzed the relationship between the two lines. We found that both lines lie in the same plane and are not parallel. This implies that they must intersect. We confirmed this by calculating the scalar triple product using the shortest distance formula, which resulted in zero. Therefore, the shortest distance between the two lines is 0.

5. Final Answer

The final answer is $\boxed{0}$, which corresponds to option (C).