To find the shortest distance between the given lines, we will follow a systematic approach using the formula for the shortest distance between two skew lines.
### Step-by-Step Solution
#### Part (i)
1. **Identify the vector equations of the lines:**
- Line 1: \(\vec{r_1} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2t) \hat{k}\)
- Line 2: \(\vec{r_2} = (s + 1) \hat{i} + (2s - 1) \hat{j} - (2s + 1) \hat{k}\)
2. **Convert the vector equations into standard form:**
- For Line 1:
- Point \(A = (1, -2, 3)\) (when \(t=0\))
- Direction vector \(B = (-1, 1, -2)\) (coefficients of \(t\))
- For Line 2:
- Point \(C = (1, -1, -1)\) (when \(s=0\))
- Direction vector \(D = (1, 2, -2)\) (coefficients of \(s\))
3. **Calculate \( \vec{C} - \vec{A} \):**
\[
\vec{C} - \vec{A} = (1 - 1) \hat{i} + (-1 + 2) \hat{j} + (-1 - 3) \hat{k} = (0) \hat{i} + (1) \hat{j} + (-4) \hat{k} = \hat{j} - 4\hat{k}
\]
4. **Calculate the cross product \( \vec{B} \times \vec{D} \):**
\[
\vec{B} = (-1, 1, -2), \quad \vec{D} = (1, 2, -2)
\]
\[
\vec{B} \times \vec{D} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 1 & -2 \\
1 & 2 & -2
\end{vmatrix}
\]
\[
= \hat{i} \left(1 \cdot (-2) - (-2) \cdot 2\right) - \hat{j} \left((-1) \cdot (-2) - (-2) \cdot 1\right) + \hat{k} \left((-1) \cdot 2 - 1 \cdot 1\right)
\]
\[
= \hat{i}(-2 + 4) - \hat{j}(2 - 2) + \hat{k}(-2 - 1) = 2\hat{i} + 0\hat{j} - 3\hat{k} = 2\hat{i} - 3\hat{k}
\]
5. **Calculate the magnitude of \( \vec{B} \times \vec{D} \):**
\[
|\vec{B} \times \vec{D}| = \sqrt{2^2 + 0^2 + (-3)^2} = \sqrt{4 + 0 + 9} = \sqrt{13}
\]
6. **Calculate the dot product \( (\vec{C} - \vec{A}) \cdot (\vec{B} \times \vec{D}) \):**
\[
(\vec{C} - \vec{A}) \cdot (\vec{B} \times \vec{D}) = (0, 1, -4) \cdot (2, 0, -3) = 0 \cdot 2 + 1 \cdot 0 + (-4) \cdot (-3) = 12
\]
7. **Calculate the shortest distance \(D\):**
\[
D = \frac{|(\vec{C} - \vec{A}) \cdot (\vec{B} \times \vec{D})|}{|\vec{B} \times \vec{D}|} = \frac{12}{\sqrt{13}}
\]
#### Part (ii)
1. **Identify the vector equations of the lines:**
- Line 1: \(\vec{r_1} = (3 - t) \hat{i} + (4 + 2t) \hat{j} + (t - 2) \hat{k}\)
- Line 2: \(\vec{r_2} = (1 + s) \hat{i} + (3s - 7) \hat{j} + (2s - 2) \hat{k}\)
2. **Convert the vector equations into standard form:**
- For Line 1:
- Point \(A = (3, 4, -2)\) (when \(t=0\))
- Direction vector \(B = (-1, 2, 1)\) (coefficients of \(t\))
- For Line 2:
- Point \(C = (1, -7, -2)\) (when \(s=0\))
- Direction vector \(D = (1, 3, 2)\) (coefficients of \(s\))
3. **Calculate \( \vec{C} - \vec{A} \):**
\[
\vec{C} - \vec{A} = (1 - 3) \hat{i} + (-7 - 4) \hat{j} + (-2 + 2) \hat{k} = (-2) \hat{i} + (-11) \hat{j} + (0) \hat{k} = -2\hat{i} - 11\hat{j}
\]
4. **Calculate the cross product \( \vec{B} \times \vec{D} \):**
\[
\vec{B} = (-1, 2, 1), \quad \vec{D} = (1, 3, 2)
\]
\[
\vec{B} \times \vec{D} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 2 & 1 \\
1 & 3 & 2
\end{vmatrix}
\]
\[
= \hat{i} \left(2 \cdot 2 - 1 \cdot 3\right) - \hat{j} \left((-1) \cdot 2 - 1 \cdot 1\right) + \hat{k} \left((-1) \cdot 3 - 2 \cdot 1\right)
\]
\[
= \hat{i}(4 - 3) - \hat{j}(-2 - 1) + \hat{k}(-3 - 2) = 1\hat{i} + 3\hat{j} - 5\hat{k}
\]
5. **Calculate the magnitude of \( \vec{B} \times \vec{D} \):**
\[
|\vec{B} \times \vec{D}| = \sqrt{1^2 + 3^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}
\]
6. **Calculate the dot product \( (\vec{C} - \vec{A}) \cdot (\vec{B} \times \vec{D}) \):**
\[
(\vec{C} - \vec{A}) \cdot (\vec{B} \times \vec{D}) = (-2, -11, 0) \cdot (1, 3, -5) = (-2) \cdot 1 + (-11) \cdot 3 + 0 \cdot (-5) = -2 - 33 + 0 = -35
\]
7. **Calculate the shortest distance \(D\):**
\[
D = \frac{|(\vec{C} - \vec{A}) \cdot (\vec{B} \times \vec{D})|}{|\vec{B} \times \vec{D}|} = \frac{35}{\sqrt{35}} = \sqrt{35}
\]
### Final Answers
- Shortest distance between the first pair of lines: \(\frac{12}{\sqrt{13}}\)
- Shortest distance between the second pair of lines: \(\sqrt{35}\)
