One vertex of a rectangular parallelopiped is at the origin O and...

Given,

One vertex of a rectangular parallelopiped is at the origin $O$ and the lengths of its edges along $x, y$ and $z$ axes are $3, 4$ and $5$ units respectively,

And $P$ be the vertex $( 3, 4, 5 )$,

Now plotting the diagram we get,

Now equation of line $OP: \dfrac{x}{3} = \dfrac{y}{4} = \dfrac{z}{5}$

And equation of line $AB: \dfrac{x - 3}{0} = \dfrac{y}{0} = \dfrac{z}{1}$

Now finding, \overset{\to}{{n}_{1}} \times \overset{\to}{{n}_{2}} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 3 & 4 & 5 \\ 0 & 0 & 1\end{cases}\right|

⇒ \overset{\to}{{n}_{1}} \times \overset{\to}{{n}_{2}} = \overset{^}{i}\left(4\right) - \overset{^}{j}\left(3\right) + \overset{^}{k}\left(0\right)

⇒ \overset{\to}{{n}_{1}} \times \overset{\to}{{n}_{2}} = 4\overset{^}{i} - 3\overset{^}{j}

Now we know that,

Shortest Distance, $S . D = \dfrac{\left(\overset{\to}{{a}_{2}} - \overset{\to}{{a}_{1}}\right) \cdot \left(\overset{\to}{{n}_{1}} \times \overset{\to}{{n}_{2}}\right)}{\left|\overset{\to}{{n}_{1}} \times \overset{\to}{{n}_{2}}\right|}$

⇒ S . D = \dfrac{\left(3\overset{^}{i} + 0\overset{^}{j} + 0\overset{^}{k} - 0\overset{^}{i} + 0\overset{^}{j} + 0\overset{^}{k}\right) \cdot \left(4\overset{^}{i} - 3\overset{^}{j}\right)}{5}

⇒ S . D = \dfrac{\left(3\overset{^}{i}\right) \cdot \left(4\overset{^}{i} - 3\overset{^}{j}\right)}{5}

$⇒ S . D = \dfrac{12}{5}$