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(A) Let the two lines be $L_1$ and $L_2$. The equation of line $L_1$ is $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$,where $\vec{a}_1 = \hat{i} + 2\hat{j

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$1/\sqrt{6}$

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(A) Let the two lines be $L_1$ and $L_2$. The equation of line $L_1$ is $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$,where $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.The equation of line $L_2$ is $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$,where $\vec{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{b}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}$.The shortest distance $d$ between two skew lines is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|}$.First,calculate $\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$.Next,calculate $\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$.The magnitude $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1+4+1} = \sqrt{6}$.Now,calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2) = (1)(-1) + (2)(2) + (2)(-1) = -1 + 4 - 2 = 1$.Thus,the shortest distance $d = \frac{|1|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$.

The shortest distance between the line passing through the point $\bar{i} + 2\bar{j} + 3\bar{k}$ and parallel to the vector $2\bar{i} + 3\bar{j} + 4\bar{k}$ and the line passing through the point $2\bar{i} + 4\bar{j} + 5\bar{k}$ and parallel to the vector $3\bar{i} + 4\bar{j} + 5\bar{k}$ is:

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