Consider the line
`L1=(x+1)/3=(y+2)/1=(z+1)/2`
`L2=(x-2)/1=(y+2)/2=(z-3)/3`
The shortest distance between `L_1` and `L_2` is

Options

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

Answer

To find the shortest distance between the two lines \( L_1 \) and \( L_2 \), we can use the formula for the shortest distance between two skew lines in 3D space. The formula is given by:

\[ S = \frac{|\mathbf{B} - \mathbf{A} \cdot (\mathbf{C} \times \mathbf{D})|}{|\mathbf{C} \times \mathbf{D}|} \]

where:
- \( \mathbf{B} \) is a point on line \( L_1 \)
- \( \mathbf{A} \) is a point on line \( L_2 \)
- \( \mathbf{C} \) is the direction vector of line \( L_1 \)
- \( \mathbf{D} \) is the direction vector of line \( L_2 \)

### Step 1: Identify points and direction vectors

From the equations of the lines, we can identify:
- For line \( L_1 \):
- A point \( \mathbf{B} = (-1, -2, -1) \)
- Direction vector \( \mathbf{C} = (3, 1, 2) \)

- For line \( L_2 \):
- A point \( \mathbf{A} = (2, -2, 3) \)
- Direction vector \( \mathbf{D} = (1, 2, 3) \)

### Step 2: Calculate \( \mathbf{B} - \mathbf{A} \)

Now, we compute \( \mathbf{B} - \mathbf{A} \):

\[ \mathbf{B} - \mathbf{A} = (-1, -2, -1) - (2, -2, 3) = (-1 - 2, -2 - (-2), -1 - 3) = (-3, 0, -4) \]

### Step 3: Calculate \( \mathbf{C} \times \mathbf{D} \)

Next, we calculate the cross product \( \mathbf{C} \times \mathbf{D} \):

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

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 modulus of \( \mathbf{C} \times \mathbf{D} \)

Now we find the magnitude of \( \mathbf{C} \times \mathbf{D} \):

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

### Step 5: Calculate the dot product \( (\mathbf{B} - \mathbf{A}) \cdot (\mathbf{C} \times \mathbf{D}) \)

Now we compute the dot product:

\[ (\mathbf{B} - \mathbf{A}) \cdot (\mathbf{C} \times \mathbf{D}) = (-3, 0, -4) \cdot (-1, -7, 5) \]
\[ = (-3)(-1) + (0)(-7) + (-4)(5) = 3 + 0 - 20 = -17 \]

### Step 6: Substitute into the formula for \( S \)

Now we substitute everything into the formula for \( S \):

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

### Final Answer

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

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