Detailed Solution<\/h3>\r\n <\/div>\r\n\r\n

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Step 1: Write the equation of line $L_1$

The line $L_1$ passes through the point (2, -1, 1) and is parallel to the vector $\vec{b_1} = \hat{i} + \hat{j} + 3\hat{k}$. Therefore, the equation of $L_1$ is given by: $\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + 3\hat{k})$

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Step 2: Write the equation of line $L_2$

The equation of line $L_2$ is given as: $\vec{r} = \hat{i} + (2\mu+1)\hat{j} - (\mu+2)\hat{k} = (\hat{i} + \hat{j} - 2\hat{k}) + \mu(2\hat{j} - \hat{k})$. Thus, $L_2$ passes through the point (1, 1, -2) and is parallel to the vector $\vec{b_2} = 2\hat{j} - \hat{k}$.

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Step 3: Find the vector connecting a point on $L_1$ to a point on $L_2$

Let $\vec{a_1} = 2\hat{i} - \hat{j} + \hat{k}$ be a point on $L_1$ and $\vec{a_2} = \hat{i} + \hat{j} - 2\hat{k}$ be a point on $L_2$. Then, the vector connecting these two points is: $\vec{a_2} - \vec{a_1} = (\hat{i} + \hat{j} - 2\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} + 2\hat{j} - 3\hat{k}$.

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Step 4: Calculate the cross product of the direction vectors

The direction vectors of the lines are $\vec{b_1} = \hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b_2} = 2\hat{j} - \hat{k}$. Their cross product is: $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 3 \\ 0 & 2 & -1 \end{vmatrix} = (-1 - 6)\hat{i} - (-1 - 0)\hat{j} + (2 - 0)\hat{k} = -7\hat{i} + \hat{j} + 2\hat{k}$.

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Step 5: Calculate the magnitude of the cross product

The magnitude of the cross product is: $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-7)^2 + 1^2 + 2^2} = \sqrt{49 + 1 + 4} = \sqrt{54} = 3\sqrt{6}$.

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Step 6: Calculate the shortest distance

The shortest distance between the lines is given by: $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$. We have $(\vec{a_2} - \vec{a_1}) = -\hat{i} + 2\hat{j} - 3\hat{k}$ and $(\vec{b_1} \times \vec{b_2}) = -7\hat{i} + \hat{j} + 2\hat{k}$. Therefore, $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-1)(-7) + (2)(1) + (-3)(2) = 7 + 2 - 6 = 3$. Thus, the shortest distance is: $d = \frac{|3|}{3\sqrt{6}} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6}$.

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\r\n Final Answer: $\frac{\sqrt{6}}{6}$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>

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