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(B) The lines are $L_1: \frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $L_2: \frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$.Passing points

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$2\sqrt{3}$

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(B) The lines are $L_1: \frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $L_2: \frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$.Passing points are $A = (\lambda, 2, 1)$ and $B = (\sqrt{3}, 1, 2)$.Direction vectors are $\vec{v_1} = -2\hat{i} + \hat{j} + \hat{k}$ and $\vec{v_2} = \hat{i} - 2\hat{j} + \hat{k}$.The cross product $\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 1 & 1 \ 1 & -2 & 1 \end{vmatrix} = \hat{i}(1+2) - \hat{j}(-2-1) + \hat{k}(4-1) = 3\hat{i} + 3\hat{j} + 3\hat{k}$.The magnitude is $|\vec{n}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$.The shortest distance $d$ is given by $d = \frac{|(\vec{B}-\vec{A}) \cdot \vec{n}|}{|\vec{n}|}$.$\vec{B}-\vec{A} = (\sqrt{3}-\lambda)\hat{i} - \hat{j} + \hat{k}$.$d = \frac{|3(\sqrt{3}-\lambda) - 3 + 3|}{3\sqrt{3}} = \frac{|3(\sqrt{3}-\lambda)|}{3\sqrt{3}} = \frac{|\sqrt{3}-\lambda|}{\sqrt{3}}$.Given $d = 1$,so $|\sqrt{3}-\lambda| = \sqrt{3}$.This implies $\sqrt{3}-\lambda = \sqrt{3}$ or $\sqrt{3}-\lambda = -\sqrt{3}$.Thus,$\lambda = 0$ or $\lambda = 2\sqrt{3}$.The sum of all possible values of $\lambda$ is $0 + 2\sqrt{3} = 2\sqrt{3}$.

If the shortest distance between the lines $\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is $1$,then the sum of all possible values of $\lambda$ is:

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