Determine whether or not the following pairs of lines intersect :
`vec(r) = (hati - 2 hatj + 3 hatk) + lambda (- hati + hatj - 2 hatk)`
and `vec(r) = ( hati - hatj - hatk) + mu (hati + 2 hatj - 2 hatk)`.

Answer

To determine whether the two given lines intersect, we can use the method of finding the shortest distance between the two lines. If the shortest distance is zero, then the lines intersect; otherwise, they do not.

### Step-by-step Solution:

1. **Identify the lines in vector form:**
The first line is given by:
\[ \vec{r_1} = (\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda (-\hat{i} + \hat{j} - 2\hat{k}) \]
The second line is given by:
\[ \vec{r_2} = (\hat{i} - \hat{j} - \hat{k}) + \mu (\hat{i} + 2\hat{j} - 2\hat{k}) \]

2. **Extract points and direction vectors:**
From the equations, we can identify:
- For the first line:
- Point \( \vec{a} = (1, -2, 3) \)
- Direction vector \( \vec{b} = (-1, 1, -2) \)
- For the second line:
- Point \( \vec{c} = (1, -1, -1) \)
- Direction vector \( \vec{d} = (1, 2, -2) \)

3. **Calculate the vector \( \vec{c} - \vec{a} \):**
\[ \vec{c} - \vec{a} = (1 - 1, -1 + 2, -1 - 3) = (0, 1, -4) \]

4. **Calculate the cross product \( \vec{b} \times \vec{d} \):**
\[ \vec{b} = (-1, 1, -2), \quad \vec{d} = (1, 2, -2) \]
Using the determinant to find the cross product:
\[ \vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i} \begin{vmatrix} 1 & -2 \\ 2 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & -2 \\ 1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} \]

Calculating the determinants:
\[ = \hat{i} (1 \cdot -2 - (-2) \cdot 2) - \hat{j} (-1 \cdot -2 - (-2) \cdot 1) + \hat{k} (-1 \cdot 2 - 1 \cdot 1) \]
\[ = \hat{i} (-2 + 4) - \hat{j} (2 - 2) + \hat{k} (-2 - 1) \]
\[ = 2\hat{i} + 0\hat{j} - 3\hat{k} = (2, 0, -3) \]

5. **Calculate the dot product \( (\vec{c} - \vec{a}) \cdot (\vec{b} \times \vec{d}) \):**
\[ \vec{c} - \vec{a} = (0, 1, -4), \quad \vec{b} \times \vec{d} = (2, 0, -3) \]
\[ (0, 1, -4) \cdot (2, 0, -3) = 0 \cdot 2 + 1 \cdot 0 + (-4) \cdot (-3) = 0 + 0 + 12 = 12 \]

6. **Conclusion:**
Since the dot product is not zero (\(12 \neq 0\)), the lines do not intersect.

### Final Answer:
The two lines do not intersect.