The line L given by frac{x - 2}{2} = frac{y - | vedclass

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(B) The line $L$ passes through $(1, 2, 3)$,so $\frac{1 - 2}{2} = \frac{2 - 1}{b} = \frac{3 + 1}{c} \Rightarrow -\frac{1}{2} = \frac{1}{b} = \frac{4}{

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$\frac{\sqrt{243}}{3}$

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(B) The line $L$ passes through $(1, 2, 3)$,so $\frac{1 - 2}{2} = \frac{2 - 1}{b} = \frac{3 + 1}{c} \Rightarrow -\frac{1}{2} = \frac{1}{b} = \frac{4}{c}$.Thus,$b = -2$ and $c = -8$.The line $L$ has direction vector $\vec{v} = (2, -2, -8)$.Since line $K$ is parallel to $L$,its direction vector must be proportional to $(2, -2, -8)$.Given $K: \frac{x + 2}{a} = \frac{y - 3}{2} = \frac{z + 4}{d}$,we have $\frac{a}{2} = \frac{2}{-2} = \frac{d}{-8} \Rightarrow \frac{a}{2} = -1 = \frac{d}{-8}$.Thus,$a = -2$ and $d = 8$.Line $L$ passes through $P_1 = (2, 1, -1)$ with direction $\vec{v} = (2, -2, -8)$.Line $K$ passes through $P_2 = (-2, 3, -4)$ with direction $\vec{v} = (2, -2, -8)$.The distance $d$ between parallel lines is given by $d = \frac{|\vec{P_1P_2} 	imes \vec{v}|}{|\vec{v}|}$.$\vec{P_1P_2} = (-2 - 2, 3 - 1, -4 - (-1)) = (-4, 2, -3)$.$\vec{P_1P_2} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -4 & 2 & -3 \ 2 & -2 & -8 \end{vmatrix} = \hat{i}(-16 - 6) - \hat{j}(32 + 6) + \hat{k}(8 - 4) = (-22, -38, 4)$.$|\vec{P_1P_2} 	imes \vec{v}| = \sqrt{(-22)^2 + (-38)^2 + 4^2} = \sqrt{484 + 1444 + 16} = \sqrt{1944}$.$|\vec{v}| = \sqrt{2^2 + (-2)^2 + (-8)^2} = \sqrt{4 + 4 + 64} = \sqrt{72}$.Distance $= \frac{\sqrt{1944}}{\sqrt{72}} = \sqrt{\frac{1944}{72}} = \sqrt{27} = 3\sqrt{3} = \sqrt{27} = \frac{\sqrt{243}}{3}$.

The line $L$ given by $\frac{x - 2}{2} = \frac{y - 1}{b} = \frac{z + 1}{c}$ passes through the point $(1, 2, 3)$. Another line $K$ is parallel to line $L$ and has the equation $\frac{x + 2}{a} = \frac{y - 3}{2} = \frac{z + 4}{d}$. Then the distance between line $L$ and $K$ is

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