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Step-by-Step Detailed Solution
Step 1: Express Each Line in Vector Form
The first line is given by
$L_1: \frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+4}{2}.$
Let the parameter for this line be $t.$
Then a point on $L_1$ can be written as
$$(x, y, z) = (1 + 2t,\; -1 - 3t,\; -4 + 2t).$$
Therefore, the direction vector of $L_1$ is
$$\vec{d_1} = \langle 2, -3, 2 \rangle.$$
The second line $L_2$ is described as passing through points
$A(-4,4,3)$ and $B(-1,6,3)$,
and it is perpendicular to the line
$$\frac{x - 3}{-2} = \frac{y}{3} = \frac{z - 1}{1},$$
whose direction vector is
$$\langle -2, 3, 1\rangle.$$
Step 2: Verify the Direction Vector of $L_2$
The vector
$$\vec{AB} = B - A = \langle -1 - (-4),\; 6 - 4,\; 3 - 3\rangle = \langle 3, 2, 0\rangle.$$
We check the dot product with $\langle -2, 3, 1\rangle$:
$$(3)(-2) + (2)(3) + (0)(1) = -6 + 6 + 0 = 0.$$
Since the dot product is zero, $\vec{AB}$ is perpendicular to $\langle -2, 3, 1\rangle.$
Thus, the direction vector of $L_2$ can be taken as
$$\vec{d_2} = \langle 3, 2, 0\rangle.$$
Hence, $L_2$ in a convenient form is:
$$
L_2:\quad \frac{x+4}{3} = \frac{y-4}{2} = \frac{z-3}{0}.
$$
Step 3: Pick Convenient Points on Each Line
A convenient point on $L_1$ can be found by setting the parameter $t = 0$,
giving the point
$$P_1(1, -1, -4).$$
A convenient point on $L_2$ is already known: $A(-4, 4, 3).$
Step 4: Form the Vector Connecting These Points
Consider the vector
$$
\overrightarrow{P_1A} = A - P_1
= \langle -4 - 1,\; 4 - (-1),\; 3 - (-4)\rangle
= \langle -5,\; 5,\; 7\rangle.
$$
Step 5: Use the Formula for the Distance Between Two Skew Lines
The shortest distance $D$ between two skew lines, with direction vectors
$\vec{d_1}$ and $\vec{d_2}$ and a connecting vector $\overrightarrow{r} = \overrightarrow{P_1A}$,
is given by:
$$
D = \frac{\bigl|\overrightarrow{P_1A} \cdot \bigl(\vec{d_1} \times \vec{d_2}\bigr)\bigr|}
{\bigl|\vec{d_1} \times \vec{d_2}\bigr|}.
$$
Step 6: Compute the Scalar Triple Product (Numerator)
We form the determinant for the scalar triple product:
$$
\begin{vmatrix}
-5 & 5 & 7 \\
2 & -3 & 2 \\
3 & 2 & 0
\end{vmatrix}.
$$
Expanding along the first row:
$$
= (-5)\begin{vmatrix}-3 & 2 \\ 2 & 0\end{vmatrix}
\;-\; 5\begin{vmatrix}2 & 2 \\ 3 & 0\end{vmatrix}
\;+\; 7\begin{vmatrix}2 & -3 \\ 3 & 2\end{vmatrix}.
$$
1. $\begin{vmatrix}-3 & 2 \\ 2 & 0\end{vmatrix} = (-3)(0) - (2)(2) = -4.$
2. $\begin{vmatrix}2 & 2 \\ 3 & 0\end{vmatrix} = (2)(0) - (2)(3) = -6.$
3. $\begin{vmatrix}2 & -3 \\ 3 & 2\end{vmatrix} = (2)(2) - (-3)(3) = 4 + 9 = 13.$
Substituting these values:
$$
= (-5)(-4) - 5(-6) + 7(13)
= 20 + 30 + 91
= 141.
$$
The absolute value of this scalar triple product is $141.$
Step 7: Compute the Magnitude of the Cross Product (Denominator)
Next, compute $\vec{d_1} \times \vec{d_2}$, where
$\vec{d_1} = \langle 2, -3, 2\rangle$ and
$\vec{d_2} = \langle 3, 2, 0\rangle.$
$$
\vec{d_1} \times \vec{d_2}
= \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & -3 & 2 \\
3 & 2 & 0
\end{vmatrix}.
$$
Expanding:
$$
= \mathbf{i}\,((-3)(0) - (2)(2))
\;-\;\mathbf{j}\,((2)(0) - (2)(3))
\;+\;\mathbf{k}\,((2)(2) - (-3)(3)).
$$
$$
= \mathbf{i}\,(-4)
\;-\;\mathbf{j}\,(-6)
\;+\;\mathbf{k}\,(4 + 9).
$$
$$
= -4\,\mathbf{i} + 6\,\mathbf{j} + 13\,\mathbf{k}.
$$
Its magnitude is
$$
\bigl|\vec{d_1} \times \vec{d_2}\bigr|
= \sqrt{(-4)^2 + 6^2 + 13^2}
= \sqrt{16 + 36 + 169}
= \sqrt{221}.
$$
Step 8: Calculate the Required Distance
Therefore, the shortest distance $D$ between the two lines $L_1$ and $L_2$ is:
$$
D = \frac{|\overrightarrow{P_1A} \cdot (\vec{d_1} \times \vec{d_2})|}
{|\vec{d_1} \times \vec{d_2}|}
= \frac{141}{\sqrt{221}}.
$$
This matches the given correct option.
Final Answer: $\frac{141}{\sqrt{221}}$
