The shortest distance between the lines (x - 4)/(4) = (y + 2)/(5) =...

Given:

$\dfrac{x - 4}{4} = \dfrac{y + 2}{5} = \dfrac{z + 3}{3}$

$\dfrac{x - 1}{3} = \dfrac{y - 3}{4} = \dfrac{z - 4}{2}$

So,

\overset{\to}{a}1 = 4\overset{^}{i} - 2\overset{^}{j} - 3\overset{^}{k}\overset{\to}{a}2 = \overset{^}{i} + 3\overset{^}{j} + 4\overset{^}{k}

So,

\overset{\to}{a}2 - \overset{\to}{a}1 = - 3\overset{^}{i} + 5\overset{^}{j} + 7\overset{^}{k}

And,

\overset{\to}{b}1 \times \overset{\to}{b}2 = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2\end{cases}\right| = - 2\overset{^}{i} + \overset{^}{j} + \overset{^}{k}

$⇒ \left|\overset{\to}{b}1 \times \overset{\to}{b}2\right| = \sqrt{{\left(- 2\right)}^{2} + {\left(1\right)}^{2} + {\left(1\right)}^{2}} = \sqrt{6}$

Shortest distance between the lines

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

= \left|\dfrac{\left(- 3\overset{^}{i} + 5\overset{^}{j} + 7\overset{^}{k}\right) \cdot \left(- 2\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)}{\sqrt{6}}\right|

$= \left|\dfrac{6 + 5 + 7}{\sqrt{6}}\right| = \dfrac{6 \times 3}{\sqrt{6}}units$

$= 3\sqrt{6}units$

Hence this is the correct option.