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(A) The given lines are $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $L_2: \vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda(\hat{i}+2\hat{j})$.The

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$\frac{8}{3\sqrt{5}}$

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(A) The given lines are $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $L_2: \vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda(\hat{i}+2\hat{j})$.The point on $L_1$ is $A(-1, -2, -1)$ and its direction vector is $\vec{b_1} = 3\hat{i} + \hat{j} + 2\hat{k}$.The point on $L_2$ is $B(2, -2, 3)$ and its direction vector is $\vec{b_2} = \hat{i} + 2\hat{j} + 0\hat{k}$.The vector $\vec{AB} = (2 - (-1))\hat{i} + (-2 - (-2))\hat{j} + (3 - (-1))\hat{k} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.The cross product $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & 2 \ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0-4) - \hat{j}(0-2) + \hat{k}(6-1) = -4\hat{i} + 2\hat{j} + 5\hat{k}$.The magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-4)^2 + 2^2 + 5^2} = \sqrt{16 + 4 + 25} = \sqrt{45} = 3\sqrt{5}$.The shortest distance $d = \frac{|\vec{AB} \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|} = \frac{|(3\hat{i} + 0\hat{j} + 4\hat{k}) \cdot (-4\hat{i} + 2\hat{j} + 5\hat{k})|}{3\sqrt{5}} = \frac{|-12 + 0 + 20|}{3\sqrt{5}} = \frac{8}{3\sqrt{5}}$.

The shortest distance (in units) between the lines $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda(\hat{i}+2\hat{j})$ is

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