Let A equiv (lambda + 2, 1 - 2lambda, lambda | vedclass

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(D) The points $A$ and $B$ represent two lines in $3D$ space.Line $L_1$ passing through $A$ can be written as $\vec{r} = (2, 1, 2) + \lambda(1, -2, 1)

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$\frac{3}{\sqrt{35}}$

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(D) The points $A$ and $B$ represent two lines in $3D$ space.Line $L_1$ passing through $A$ can be written as $\vec{r} = (2, 1, 2) + \lambda(1, -2, 1)$.Line $L_2$ passing through $B$ can be written as $\vec{r} = (1, 0, 1) + k(2, 1, 1)$.The shortest distance between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + k \vec{b_2}$ is given by $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})}{|\vec{b_1} 	imes \vec{b_2}|} ight|$.Here,$\vec{a_2} - \vec{a_1} = (1-2, 0-1, 1-2) = (-1, -1, -1)$.$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 1 \ 2 & 1 & 1 \end{vmatrix} = \hat{i}(-2-1) - \hat{j}(1-2) + \hat{k}(1+4) = -3\hat{i} + \hat{j} + 5\hat{k}$.Magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-3)^2 + 1^2 + 5^2} = \sqrt{9 + 1 + 25} = \sqrt{35}$.Shortest distance $d = \left| \frac{(-1, -1, -1) \cdot (-3, 1, 5)}{\sqrt{35}} ight| = \left| \frac{3 - 1 - 5}{\sqrt{35}} ight| = \left| \frac{-3}{\sqrt{35}} ight| = \frac{3}{\sqrt{35}}$.

Let $A \equiv (\lambda + 2, 1 - 2\lambda, \lambda + 2)$ and $B \equiv (2k + 1, k, k + 1)$ where $\lambda, k \in \mathbb{R}$. Then the minimum distance between $A$ and $B$ is -

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