If $\mathrm{d}_1$ is the shortest distance between the lines $x+1=2 y=-12 z, x=y+2=6 z-6$ and $\mathrm{d}_2$ is the shortest distance between the lines $$\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$$, then the value of $\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}$ is :

<p>$$\mathrm{L}_1: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{-1 / 12}, \mathrm{~L}_2: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\frac{1}{6}}$$</p> <p>$\mathrm{d}_1=$ shortest distance between $\mathrm{L}_1 ~\& \mathrm{~L}_2$</p> <p>$$\begin{aligned} & =\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}\right| \\ & d_1=2 \\ & \mathrm{~L}_3: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+8}{-7}=\frac{\mathrm{z}-4}{5}, \mathrm{~L}_4: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-6}{-3} \\ & \mathrm{~d}_2=\text { shortest distance between } \mathrm{L}_3 ~\& \mathrm{~L}_4 \\ & \mathrm{~d}_2=\frac{12}{\sqrt{3}} \text { Hence } \\ & =\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}=16 \end{aligned}$$</p>