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(B) The direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = (3, 5, 7)$ and $\vec{v_2} = (1, 4, 7)$.Since line $L$ is perpendicular to both,its direc

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$\frac{5}{2}\sqrt{2}$

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(B) The direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = (3, 5, 7)$ and $\vec{v_2} = (1, 4, 7)$.Since line $L$ is perpendicular to both,its direction vector $\vec{v}$ is parallel to $\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 5 & 7 \ 1 & 4 & 7 \end{vmatrix} = \hat{i}(35-28) - \hat{j}(21-7) + \hat{k}(12-5) = 7\hat{i} - 14\hat{j} + 7\hat{k}$.Dividing by $7$,we take $\vec{v} = (1, -2, 1)$.The direction vector of $L_3$ is $\vec{v_3} = (2, 1, 2)$.The cosine of the angle $	heta$ between them is $\cos 	heta = \frac{|(1)(2) + (-2)(1) + (1)(2)|}{\sqrt{1^2+(-2)^2+1^2} \sqrt{2^2+1^2+2^2}} = \frac{|2-2+2|}{\sqrt{6}\sqrt{9}} = \frac{2}{3\sqrt{6}}$.Since $\cos 	heta = \frac{2}{3\sqrt{6}}$,we have $\cos^2 	heta = \frac{4}{9 \cdot 6} = \frac{4}{54} = \frac{2}{27}$.Then $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{2}{27} = \frac{25}{27}$,so $\sin 	heta = \frac{5}{3\sqrt{3}}$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{5}{3\sqrt{3}} \cdot \frac{3\sqrt{6}}{2} = \frac{5\sqrt{6}}{2\sqrt{3}} = \frac{5\sqrt{2}}{2} = \frac{5}{2}\sqrt{2}$.

Let a line $L$ be perpendicular to both the lines $L_1: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$ and $L_2: \frac{x-2}{1} = \frac{y-4}{4} = \frac{z-6}{7}$. If $\theta$ is the acute angle between the lines $L$ and $L_3: \frac{x-7}{2} = \frac{y-7}{1} = \frac{z}{2}$,then $\tan \theta$ is equal to:

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