Read the following passage and answer the questions. Consider the lines
`L_(1) : (x+1)/(3)=(y+2)/(1)=(z+1)/(2)`
`L_(2) : (x-2)/(1)=(y+2)/(2)=(z-3)/(3)`
Q. The shortest distance between `L_(1) and L_(2)` is

Options

A. `0` unit
B. `(17)/(sqrt(3))` units
C. `(41)/(5sqrt(3))` units
D. `(17)/(5sqrt(3))` units

Answer

To find the shortest distance between the two lines \( L_1 \) and \( L_2 \), we can use the formula for the distance between two skew lines in three-dimensional space. The lines are given in symmetric form, and we can extract the direction vectors and points on each line.

### Step 1: Identify the direction vectors and points on each line

For line \( L_1 : \frac{x+1}{3} = \frac{y+2}{1} = \frac{z+1}{2} \):
- Direction vector \( \mathbf{a_1} = (3, 1, 2) \)
- Point on \( L_1 \) is \( P_1(-1, -2, -1) \)

For line \( L_2 : \frac{x-2}{1} = \frac{y+2}{2} = \frac{z-3}{3} \):
- Direction vector \( \mathbf{a_2} = (1, 2, 3) \)
- Point on \( L_2 \) is \( P_2(2, -2, 3) \)

### Step 2: Use the distance formula for skew lines

The formula for the distance \( d \) between two skew lines is given by:

\[ d = \frac{|(\mathbf{P_2} - \mathbf{P_1}) \cdot (\mathbf{a_1} \times \mathbf{a_2})|}{|\mathbf{a_1} \times \mathbf{a_2}|} \]

Where:
- \( \mathbf{P_2} - \mathbf{P_1} = (2 - (-1), -2 - (-2), 3 - (-1)) = (3, 0, 4) \)

### Step 3: Calculate the cross product \( \mathbf{a_1} \times \mathbf{a_2} \)

\[ \mathbf{a_1} \times \mathbf{a_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} \]

Calculating the determinant:

\[ = \mathbf{i}(1 \cdot 3 - 2 \cdot 2) - \mathbf{j}(3 \cdot 3 - 2 \cdot 1) + \mathbf{k}(3 \cdot 2 - 1 \cdot 1) \]
\[ = \mathbf{i}(3 - 4) - \mathbf{j}(9 - 2) + \mathbf{k}(6 - 1) \]
\[ = -\mathbf{i} - 7\mathbf{j} + 5\mathbf{k} \]
\[ = (-1, -7, 5) \]

### Step 4: Calculate the magnitude of \( \mathbf{a_1} \times \mathbf{a_2} \)

\[ |\mathbf{a_1} \times \mathbf{a_2}| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3} \]

### Step 5: Calculate the dot product \( (\mathbf{P_2} - \mathbf{P_1}) \cdot (\mathbf{a_1} \times \mathbf{a_2}) \)

\[ (3, 0, 4) \cdot (-1, -7, 5) = 3 \cdot (-1) + 0 \cdot (-7) + 4 \cdot 5 = -3 + 0 + 20 = 17 \]

### Step 6: Substitute into the distance formula

\[ d = \frac{|17|}{5\sqrt{3}} = \frac{17}{5\sqrt{3}} \]

### Final Answer

The shortest distance between the lines \( L_1 \) and \( L_2 \) is:

\[ \boxed{\frac{17}{5\sqrt{3}}} \]