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(B) The shortest distance $d$ between two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by $d = \

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$2$

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(B) The shortest distance $d$ between two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|}$.Given $\vec{a}_1 = -\hat{i} + 3\hat{k}$,$\vec{b}_1 = \hat{i} - a\hat{j}$,$\vec{a}_2 = -\hat{j} + 2\hat{k}$,and $\vec{b}_2 = \hat{i} - \hat{j} + \hat{k}$.$\vec{a}_2 - \vec{a}_1 = (0 - (-1))\hat{i} + (-1 - 0)\hat{j} + (2 - 3)\hat{k} = \hat{i} - \hat{j} - \hat{k}$.$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -a & 0 \ 1 & -1 & 1 \end{vmatrix} = \hat{i}(-a) - \hat{j}(1) + \hat{k}(-1 + a) = -a\hat{i} - \hat{j} + (a-1)\hat{k}$.$|\vec{b}_1 	imes \vec{b}_2| = \sqrt{(-a)^2 + (-1)^2 + (a-1)^2} = \sqrt{a^2 + 1 + a^2 - 2a + 1} = \sqrt{2a^2 - 2a + 2}$.$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2) = (1)(-a) + (-1)(-1) + (-1)(a-1) = -a + 1 - a + 1 = 2 - 2a$.Given $d = \sqrt{\frac{2}{3}}$,so $\frac{|2 - 2a|}{\sqrt{2a^2 - 2a + 2}} = \sqrt{\frac{2}{3}}$.Squaring both sides: $\frac{4(1-a)^2}{2(a^2 - a + 1)} = \frac{2}{3} \implies \frac{2(1-2a+a^2)}{a^2-a+1} = \frac{2}{3} \implies 3(a^2 - 2a + 1) = a^2 - a + 1$.$3a^2 - 6a + 3 = a^2 - a + 1 \implies 2a^2 - 5a + 2 = 0$.$(2a - 1)(a - 2) = 0$,so $a = 2$ or $a = 0.5$.The integral value of $a$ is $2$.

If the shortest distance between the lines $\vec{r}=(-\hat{i}+3\hat{k})+\lambda(\hat{i}-a\hat{j})$ and $\vec{r}=(-\hat{j}+2\hat{k})+\mu(\hat{i}-\hat{j}+\hat{k})$ is $\sqrt{\frac{2}{3}}$,then the integral value of $a$ is equal to

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