1. Identify the required edge

A rectangular parallelepiped has vertices with coordinates formed from $x\in\{0,3\},\quad y\in\{0,4\},\quad z\in\{0,5\}.$

The space diagonal is from $O=(0,0,0) \quad \text{to} \quad P=(3,4,5).$

We need an edge parallel to the $z$-axis and not passing through $O$ or $P$.

Edges parallel to the $z$-axis are:

• through $(0,0)$ in the $xy$-plane: passes through $O$,
• through $(3,4)$ in the $xy$-plane: passes through $P$,
• through $(3,0)$ in the $xy$-plane,
• through $(0,4)$ in the $xy$-plane.

So the relevant edges are: $L_1: (3,0,z), \qquad L_2: (0,4,z), \qquad 0\le z\le 5.$ By symmetry, the shortest distance from diagonal $OP$ to either of these will be the same. We compute for $L_1$.

2. Write equations of the two lines

The diagonal $OP$ has direction vector $\vec d_1=(3,4,5).$ So its parametric form is $\vec r=(0,0,0)+t(3,4,5).$

The edge $L_1$ passes through $(3,0,0)$ and is parallel to the $z$-axis, so its direction vector is $\vec d_2=(0,0,1).$ Its parametric form is $\vec r=(3,0,0)+s(0,0,1).$

3. Use formula for distance between two skew lines

For lines $\vec r=\vec a_1+t\vec d_1, \qquad \vec r=\vec a_2+s\vec d_2,$ the shortest distance is $D=\frac{|(\vec a_2-\vec a_1)\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Here, $\vec a_1=(0,0,0), \qquad \vec a_2=(3,0,0).$ Thus, $\vec a_2-\vec a_1=(3,0,0).$

Now compute the cross product: $\vec d_1\times \vec d_2=(3,4,5)\times(0,0,1).$

Using determinant form,

So, $|\vec d_1\times \vec d_2|=\sqrt{4^2+(-3)^2}=5.$

Also,

Therefore, $D=\frac{|12|}{5}=\frac{12}{5}.$

4. Check with the other possible edge

For the edge through $(0,4,0)$ parallel to $z$-axis, $\vec a_2=(0,4,0).$ Then $(0,4,0)\cdot(4,-3,0)=-12,$ so distance is again $\frac{12}{5}.$

Hence the required shortest distance is $\boxed{\frac{12}{5}}.$

5. Option check

• A: $\frac{12}{\sqrt{5}}$ — incorrect
• B: $12\sqrt{5}$ — incorrect
• C: $\frac{12}{5}$ — correct
• D: $\frac{12}{5\sqrt{5}}$ — incorrect

So the correct option is $\boxed{\text{C}}.$