1. **State the problem:** Find the shortest distance between the two lines given by their vector equations: $$\vec{r_1} = (1 - t)\hat{i} + (t - 2)\hat{j} + (3 - 2t)\hat{k}$$ $$\vec{r_2} = (s + 1)\hat{i} + (2s - 1)\hat{j} - (2s + 1)\hat{k}$$ 2. **Recall the formula:** The shortest distance $d$ between two skew lines can be found using the formula: $$d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$$ where $\vec{a_1}$ and $\vec{a_2}$ are position vectors of points on each line, and $\vec{b_1}$ and $\vec{b_2}$ are direction vectors of the lines. 3. **Identify vectors:** From line 1, when $t=0$, point $\vec{a_1} = 1\hat{i} - 2\hat{j} + 3\hat{k}$. Direction vector of line 1: $$\vec{b_1} = \frac{d}{dt}[(1 - t), (t - 2), (3 - 2t)] = (-1, 1, -2)$$ From line 2, when $s=0$, point $\vec{a_2} = 1\hat{i} - 1\hat{j} - 1\hat{k}$. Direction vector of line 2: $$\vec{b_2} = \frac{d}{ds}[(s + 1), (2s - 1), (-2s - 1)] = (1, 2, -2)$$ 4. **Calculate $\vec{a_2} - \vec{a_1}$:** $$\vec{a_2} - \vec{a_1} = (1 - 1, -1 - (-2), -1 - 3) = (0, 1, -4)$$ 5. **Calculate cross product $\vec{b_1} \times \vec{b_2}$:** $$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(1 \cdot (-2) - (-2) \cdot 2) - \hat{j}(-1 \cdot (-2) - (-2) \cdot 1) + \hat{k}(-1 \cdot 2 - 1 \cdot 1)$$ $$= \hat{i}(-2 + 4) - \hat{j}(2 + 2) + \hat{k}(-2 - 1) = 2\hat{i} - 4\hat{j} - 3\hat{k}$$ 6. **Calculate numerator $|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|$:** $$= |(0, 1, -4) \cdot (2, -4, -3)| = |0 \cdot 2 + 1 \cdot (-4) + (-4) \cdot (-3)| = |-4 + 12| = 8$$ 7. **Calculate denominator $|\vec{b_1} \times \vec{b_2}|$:** $$= \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$$ 8. **Calculate shortest distance $d$:** $$d = \frac{8}{\sqrt{29}}$$ **Final answer:** The shortest distance between the two lines is $$\boxed{\frac{8}{\sqrt{29}}}$$.