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(D) 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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$6$

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(D) 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 $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|}$.Here,$\vec{a}_1 = \alpha \hat{i} + 2 \hat{j} + 2 \hat{k}$,$\vec{b}_1 = \hat{i} - 2 \hat{j} + 2 \hat{k}$,$\vec{a}_2 = -4 \hat{i} - \hat{k}$,and $\vec{b}_2 = 3 \hat{i} - 2 \hat{j} - 2 \hat{k}$.First,calculate $\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 2 \ 3 & -2 & -2 \end{vmatrix} = \hat{i}(4 - (-4)) - \hat{j}(-2 - 6) + \hat{k}(-2 - (-6)) = 8 \hat{i} + 8 \hat{j} + 4 \hat{k}$.The magnitude $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$.Now,$\vec{a}_2 - \vec{a}_1 = (-4 - \alpha) \hat{i} - 2 \hat{j} - 3 \hat{k}$.The shortest distance is $9 = \frac{|((-4 - \alpha) \hat{i} - 2 \hat{j} - 3 \hat{k}) \cdot (8 \hat{i} + 8 \hat{j} + 4 \hat{k})|}{12}$.$9 = \frac{|8(-4 - \alpha) - 16 - 12|}{12} \implies 108 = |-32 - 8\alpha - 28| = |-60 - 8\alpha|$.Since $\alpha > 0$,$|-60 - 8\alpha| = 60 + 8\alpha$.$60 + 8\alpha = 108 \implies 8\alpha = 48 \implies \alpha = 6$.

If the shortest distance between the lines $\vec{r}_{1}=\alpha \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k}), \lambda \in R, \alpha>0$ and $\vec{r}_{2}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k}), \mu \in R$ is $9$,then $\alpha$ is equal to $.....$

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