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

Find $$\frac{32\sqrt{3} \, d_1}{d_2}$$ where $$d_1$$ and $$d_2$$ are the shortest distances between the given pairs of lines.

To determine $$d_1$$, consider the lines $$x+1 = 2y = -12z$$ and $$x = y+2 = 6z-6.$$ The first line can be parametrized as $$\frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12},$$ so its direction vector is $$(1,\,1/2,\,-1/12)$$ or equivalently $$(12,\,6,\,-1)$$, and it passes through the point $$(-1,\,0,\,0)$$. The second line can be written as $$\frac{x}{1} = \frac{y+2}{1} = \frac{z-1}{1/6},$$ giving the direction vector $$(1,\,1,\,1/6)$$ or equivalently $$(6,\,6,\,1)$$, and passing through the point $$(0,\,-2,\,1)$$.

The cross product of these direction vectors is $$\vec{d_1}\times\vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 12 & 6 & -1 \\ 6 & 6 & 1 \end{vmatrix} = \hat{i}(6+6) - \hat{j}(12+6) + \hat{k}(72-36) = (12,\,-18,\,36),$$ whose magnitude is $$|\vec{d_1}\times\vec{d_2}| = \sqrt{144 + 324 + 1296} = \sqrt{1764} = 42.$$ If $$P_1$$ and $$P_2$$ are points on the first and second lines respectively, then $$\overrightarrow{P_1P_2} = (1,\,-2,\,1),$$ so $$d_1 = \frac{\bigl|(1,-2,1)\cdot(12,-18,36)\bigr|}{42} = \frac{|12 + 36 + 36|}{42} = \frac{84}{42} = 2.$$

Next, to find $$d_2$$, consider the lines $$\frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}$$ and $$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}.$$ The first of these lines passes through $$(1,\,-8,\,4)$$ with direction vector $$(2,\,-7,\,5)$$, and the second passes through $$(1,\,2,\,6)$$ with direction vector $$(2,\,1,\,-3)$$.

Their cross product is $$\vec{d_1}\times\vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21-5) - \hat{j}(-6-10) + \hat{k}(2+14) = (16,\,16,\,16),$$ so $$|\vec{d_1}\times\vec{d_2}| = 16\sqrt{3}.$$ With $$\overrightarrow{P_1P_2} = (0,\,10,\,2),$$ we obtain $$d_2 = \frac{\bigl|(0,10,2)\cdot(16,16,16)\bigr|}{16\sqrt{3}} = \frac{|0 + 160 + 32|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}.$$

Finally, substituting these values into the given expression yields $$\frac{32\sqrt{3} \cdot d_1}{d_2} = \frac{32\sqrt{3} \times 2}{4\sqrt{3}} = \frac{64\sqrt{3}}{4\sqrt{3}} = 16.$$

The correct answer is $$\boxed{16}$$.