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(D) Given that line $L_1$ passes through $\vec{a}_1 = 2 \hat{i}-\hat{k}$ and is parallel to $\vec{b}_1 = 3 \hat{i}-\hat{j}+2 \hat{k}$.Line $L_2$ passe

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$\frac{9}{\sqrt{35}}$

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(D) Given that line $L_1$ passes through $\vec{a}_1 = 2 \hat{i}-\hat{k}$ and is parallel to $\vec{b}_1 = 3 \hat{i}-\hat{j}+2 \hat{k}$.Line $L_2$ passes through $\vec{a}_2 = 2 \hat{i}+\hat{j}-3 \hat{k}$ and is parallel to $\vec{b}_2 = \hat{i}-2 \hat{j}+\hat{k}$.The shortest distance $d$ between two skew lines is given by the formula: $d = \frac{|(\vec{b}_1 	imes \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}_1 	imes \vec{b}_2|}$.First,calculate the cross product $\vec{b}_1 	imes \vec{b}_2$:$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 2 \ 1 & -2 & 1 \end{vmatrix} = \hat{i}(-1 + 4) - \hat{j}(3 - 2) + \hat{k}(-6 + 1) = 3 \hat{i} - \hat{j} - 5 \hat{k}$.The magnitude is $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{3^2 + (-1)^2 + (-5)^2} = \sqrt{9 + 1 + 25} = \sqrt{35}$.Next,calculate $\vec{a}_2 - \vec{a}_1 = (2 \hat{i} + \hat{j} - 3 \hat{k}) - (2 \hat{i} - \hat{k}) = \hat{j} - 2 \hat{k}$.Now,calculate the dot product: $(\vec{b}_1 	imes \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (3 \hat{i} - \hat{j} - 5 \hat{k}) \cdot (0 \hat{i} + 1 \hat{j} - 2 \hat{k}) = (3)(0) + (-1)(1) + (-5)(-2) = 0 - 1 + 10 = 9$.Therefore,the shortest distance is $d = \frac{|9|}{\sqrt{35}} = \frac{9}{\sqrt{35}}$.

Let $L_1$ (respectively $L_2$) be the line passing through $2 \hat{i}-\hat{k}$ (respectively $2 \hat{i}+\hat{j}-3 \hat{k}$) and parallel to $3 \hat{i}-\hat{j}+2 \hat{k}$ (respectively $\hat{i}-2 \hat{j}+\hat{k}$). Then the shortest distance between the lines $L_1$ and $L_2$ is equal to

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