If d_1 is the shortest distance between the l | vedclass

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(B) For the first pair of lines:$L_1: \frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12} \implies \vec{a}_1 = (-1, 0, 0), \vec{b}_1 = (1, 1/2, -1/12)$$L_

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$16$

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(B) For the first pair of lines:$L_1: \frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12} \implies \vec{a}_1 = (-1, 0, 0), \vec{b}_1 = (1, 1/2, -1/12)$$L_2: \frac{x}{1} = \frac{y+2}{1} = \frac{z-1}{1/6} \implies \vec{a}_2 = (0, -2, 1), \vec{b}_2 = (1, 1, 1/6)$Shortest distance $d_1 = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|} = 2$.For the second pair of lines:$L_3: \frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}, \vec{a}_3 = (1, -8, 4), \vec{b}_3 = (2, -7, 5)$$L_4: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}, \vec{a}_4 = (1, 2, 6), \vec{b}_4 = (2, 1, -3)$Shortest distance $d_2 = \frac{|(\vec{a}_4 - \vec{a}_3) \cdot (\vec{b}_3 	imes \vec{b}_4)|}{|\vec{b}_3 	imes \vec{b}_4|} = \frac{12}{\sqrt{3}}$.Calculating the final value:$\frac{32 \sqrt{3} d_1}{d_2} = \frac{32 \sqrt{3} 	imes 2}{12/\sqrt{3}} = \frac{64 \sqrt{3} 	imes \sqrt{3}}{12} = \frac{64 	imes 3}{12} = \frac{192}{12} = 16$.

If $d_1$ is the shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$,and $d_2$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$,then the value of $\frac{32 \sqrt{3} d_1}{d_2}$ is:

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