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The given lines are: Line 1: $\vec{r}=(2\hat{i}-\hat{j}+3\hat{k})+\lambda(\hat{i}-2\hat{j}+3\hat{k})$ Line 2: $\vec{r}=(\hat{i}+4\hat{k})+\mu(3\hat{i}-6\hat{j}+9\hat{k})$ Comparing with the general form $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$, we have: $\vec{a_1} = 2\hat{i}-\hat{j}+3\hat{k}$ $\vec{b_1} = \hat{i}-2\hat{j}+3\hat{k}$ $\vec{a_2} = \hat{i}+4\hat{k}$ $\vec{b_2} = 3\hat{i}-6\hat{j}+9\hat{k}$
Observe that $\vec{b_2} = 3\vec{b_1}$, which means the lines are parallel.
$\vec{a_2} - \vec{a_1} = (\hat{i}+4\hat{k}) - (2\hat{i}-\hat{j}+3\hat{k}) = -\hat{i} + \hat{j} + \hat{k}$
$(\vec{a_2} - \vec{a_1}) \times \vec{b_1} = (-\hat{i} + \hat{j} + \hat{k}) \times (\hat{i}-2\hat{j}+3\hat{k})$ $= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} = (3+2)\hat{i} - (-3-1)\hat{j} + (2-1)\hat{k} = 5\hat{i} + 4\hat{j} + \hat{k}$
$|(\vec{a_2} - \vec{a_1}) \times \vec{b_1}| = \sqrt{5^2 + 4^2 + 1^2} = \sqrt{25 + 16 + 1} = \sqrt{42}$
$|\vec{b_1}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$
The shortest distance $d$ between the parallel lines is given by: $d = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b_1}|}{|\vec{b_1}|} = \frac{\sqrt{42}}{\sqrt{14}} = \sqrt{\frac{42}{14}} = \sqrt{3}$
Final Answer: $\sqrt{3}$
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