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(C) The lines are given by $\vec{r} = \vec{a}_1 + t\vec{p}$ and $\vec{r} = \vec{a}_2 + s\vec{q}$,where $\vec{a}_1 = \lambda \hat{i} + 2 \hat{j} + \hat

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$43$

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(C) The lines are given by $\vec{r} = \vec{a}_1 + t\vec{p}$ and $\vec{r} = \vec{a}_2 + s\vec{q}$,where $\vec{a}_1 = \lambda \hat{i} + 2 \hat{j} + \hat{k}$,$\vec{p} = 3 \hat{i} - \hat{j} + \hat{k}$,$\vec{a}_2 = -2 \hat{i} - 5 \hat{j} + 4 \hat{k}$,and $\vec{q} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$.The shortest distance $d$ is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{p} 	imes \vec{q})|}{||\vec{p} 	imes \vec{q}||}$.First,calculate $\vec{p} 	imes \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 1 \ -3 & 2 & 4 \end{vmatrix} = \hat{i}(-4-2) - \hat{j}(12+3) + \hat{k}(6-3) = -6 \hat{i} - 15 \hat{j} + 3 \hat{k}$.The magnitude $||\vec{p} 	imes \vec{q}|| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.Now,$\vec{a}_2 - \vec{a}_1 = (-2-\lambda) \hat{i} - 7 \hat{j} + 3 \hat{k}$.$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{p} 	imes \vec{q}) = (-2-\lambda)(-6) + (-7)(-15) + (3)(3) = 12 + 6\lambda + 105 + 9 = 6\lambda + 126$.Given $d = \frac{44}{\sqrt{30}}$,we have $\frac{|6\lambda + 126|}{3\sqrt{30}} = \frac{44}{\sqrt{30}}$.$|6\lambda + 126| = 132$.This implies $6\lambda + 126 = 132$ or $6\lambda + 126 = -132$.Case $1$: $6\lambda = 6 \implies \lambda = 1$.Case $2$: $6\lambda = -258 \implies \lambda = -43$.The possible values of $|\lambda|$ are $|1| = 1$ and $|-43| = 43$.Thus,the largest possible value of $|\lambda|$ is $43$.

If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$,then the largest possible value of $|\lambda|$ is equal to ..........

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