If d _ 1 is the shortest distance between the lines x + 1 = 2y = -...

Let,

${L}_{1}: \dfrac{x + 1}{1} = \dfrac{y}{\dfrac{1}{2}} = \dfrac{z}{\dfrac{- 1}{12}},$

And ${L}_{2}: \dfrac{x}{1} = \dfrac{y + 2}{1} = \dfrac{z - 1}{\dfrac{1}{6}}$

Now, given ${d}_{1} =$ shortest distance between ${L}_{1}$ and ${L}_{2}$

Now using the formula of shortest distance we get,

 ${d}_{1} = \left|\dfrac{\left(\overset{\to}{a}2 - \overset{\to}{a}1\right) \cdot \left(\overset{\to}{b}1 \times \overset{\to}{b}2\right)}{\left|\left(\overset{\to}{b}1 \times \overset{\to}{b}2\right)\right|}\right|$

⇒ {d}_{1} = \left|\dfrac{\left(\overset{^}{i} - 2\overset{^}{j} + \overset{^}{k}\right) \cdot \left(\left(\overset{^}{i} + \dfrac{1}{2}\overset{^}{j} - \dfrac{1}{12}\overset{^}{k}\right) \times \left(\overset{^}{i} + \overset{^}{j} + \dfrac{1}{6}\overset{^}{k}\right)\right)}{\left|\left(\overset{^}{i} + \dfrac{1}{2}\overset{^}{j} - \dfrac{1}{12}\overset{^}{k}\right) \times \left(\overset{^}{i} + \overset{^}{j} + \dfrac{1}{6}\overset{^}{k}\right)\right|}\right|

⇒ {d}_{1} = \left|\dfrac{\left(\overset{^}{i} - 2\overset{^}{j} + \overset{^}{k}\right) \cdot \left(\overset{^}{i}6 - \overset{^}{j}4 + \overset{^}{k}2\right)}{\left|\left(\overset{^}{i}6 - \overset{^}{j}4 + \overset{^}{k}2\right)\right|}\right|

$⇒ {d}_{1} = \left|\dfrac{\dfrac{1}{6} + \dfrac{1}{2} + \dfrac{1}{2}}{\sqrt{\dfrac{49}{144}}}\right|$

$⇒ {d}_{1} = \left|\dfrac{\dfrac{7}{6}}{\dfrac{7}{12}}\right| = 2$

Now, let 

${L}_{3}: \dfrac{x - 1}{2} = \dfrac{y + 8}{- 7} = \dfrac{z - 4}{5}, {L}_{4}: \dfrac{x - 1}{2} = \dfrac{y - 2}{1} = \dfrac{z - 6}{- 3}$

Similarly finding ${d}_{2} =$ shortest distance between ${L}_{3}$ & ${L}_{4}$ we get,

{d}_{2} = \left|\dfrac{\left(10\overset{^}{j} + 2\overset{^}{k}\right) \cdot \left(\left(2\overset{^}{i} - 7\overset{^}{j} + 5\overset{^}{k}\right) \times \left(2\overset{^}{i} + \overset{^}{j} - 3\overset{^}{k}\right)\right)}{\left|\left(\left(2\overset{^}{i} - 7\overset{^}{j} + 5\overset{^}{k}\right) \times \left(2\overset{^}{i} + \overset{^}{j} - 3\overset{^}{k}\right)\right)\right|}\right|

⇒ {d}_{2} = \left|\dfrac{\left(10\overset{^}{j} + 2\overset{^}{k}\right) \cdot \left(16\overset{^}{i} + 16\overset{^}{j} + 16\overset{^}{k}\right)}{\left|\left(16\overset{^}{i} + 16\overset{^}{j} + 16\overset{^}{k}\right)\right|}\right|

$⇒ {d}_{2} = \left|\dfrac{160 + 32}{16\sqrt{3}}\right|$

$⇒ {d}_{2} = \left|\dfrac{12}{\sqrt{3}}\right|$

Hence, $\dfrac{32\sqrt{3}{d}_{1}}{{d}_{2}} = \dfrac{32\sqrt{3} \times 2}{\dfrac{12}{\sqrt{3}}} = 16$