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(B) The general point on the line $\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z}{1} = \lambda$ is $P(2\lambda+1, -3\lambda-1, \lambda)$.The distance from

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$4\sqrt{\frac{7}{5}}$

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(B) The general point on the line $\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z}{1} = \lambda$ is $P(2\lambda+1, -3\lambda-1, \lambda)$.The distance from $(1, -1, 0)$ is $\sqrt{(2\lambda)^2 + (-3\lambda)^2 + \lambda^2} = 4\sqrt{14}$.$\sqrt{4\lambda^2 + 9\lambda^2 + \lambda^2} = 4\sqrt{14} \Rightarrow \sqrt{14\lambda^2} = 4\sqrt{14} \Rightarrow |\lambda| = 4$.For the point nearer to the origin,we choose $\lambda = -4$. Thus,$P = (2(-4)+1, -3(-4)-1, -4) = (-7, 11, -4)$.The lines are $L_1: \frac{x+7}{1} = \frac{y-11}{2} = \frac{z+4}{3}$ and $L_2: \frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$.Shortest distance $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|}$.$\vec{a_2} - \vec{a_1} = (-5 - (-7))\hat{i} + (10 - 11)\hat{j} + (3 - (-4))\hat{k} = 2\hat{i} - \hat{j} + 7\hat{k}$.$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 2 & 1 & 1 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(1-6) + \hat{k}(1-4) = -\hat{i} + 5\hat{j} - 3\hat{k}$.$d = \frac{|(2)(-1) + (-1)(5) + (7)(-3)|}{\sqrt{(-1)^2 + 5^2 + (-3)^2}} = \frac{|-2 - 5 - 21|}{\sqrt{1 + 25 + 9}} = \frac{28}{\sqrt{35}} = \frac{28}{\sqrt{5}\sqrt{7}} = \frac{4 	imes 7}{\sqrt{5}\sqrt{7}} = 4\sqrt{\frac{7}{5}}$.

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