Let Q be the cube with the set of vertices $$\{(x_1, x_2, x_3) \in \mathbb{R}^3 : x_1, x_2, x_3 \in \{0, 1\}\}$$. Let F be the set of all twelve lines containing the diagonals of the six faces of the cube Q. Let S be the set of all four lines containing the main diagonals of the cube Q; for instance, the line passing through the vertices (0, 0, 0) and (1, 1, 1) is in S. For lines $$\ell_1$$ and $$\ell_2$$, let $$d(\ell_1, \ell_2)$$ denote the shortest distance between them. Then the maximum value of $$d(\ell_1, \ell_2)$$, as $$\ell_1$$ varies over F and $$\ell_2$$ varies over S, is

The cube has side-length $$1$$ and its centre is at $$\left(\tfrac12,\tfrac12,\tfrac12\right)$$. Because of the full rotational symmetry of the cube, we may, without loss of generality, fix the body-diagonal

$$\ell_2:\; (0,0,0)+t(1,1,1),\;t\in\mathbb{R}$$

as the representative of the set $$S$$. For every face of the cube choose Cartesian coordinates in the natural way; each face diagonal has a direction vector which contains exactly two entries $$\pm1$$ and one entry $$0$$.

Write the direction vector of the fixed body diagonal as $$\mathbf{s}=(1,1,1)$$ and the direction vector of a face diagonal as $$\mathbf{f}=(a,b,c),\qquad (a,b,c)\in\{\! -1,0,1\!\},\;a^2+b^2+c^2=2,\;abc=0.$$ (The zero component indicates the coordinate which is constant on that face.)

Step 1 Magnitude of the cross product $$\mathbf{s}\times\mathbf{f}$$.
Since $$|\mathbf{s}|^2=3,\;|\mathbf{f}|^2=2$$, we use $$|\mathbf{s}\times\mathbf{f}|^2=|\mathbf{s}|^2|\mathbf{f}|^2-(\mathbf{s}\!\cdot\!\mathbf{f})^2=6-(\mathbf{s}\!\cdot\!\mathbf{f})^2.$$ Because $$\mathbf{s}\!\cdot\!\mathbf{f}=a+b+c,$$ the possible values are $$0,\;\pm2.$$ Hence

$$|\mathbf{s}\times\mathbf{f}|=\begin{cases}\sqrt6 & \text{if }\mathbf{s}\!\cdot\!\mathbf{f}=0,\\[4pt] \sqrt2 & \text{if }|\mathbf{s}\!\cdot\!\mathbf{f}|=2.\end{cases}$$

Step 2 Which face diagonals are skew to $$\ell_2$$?
If $$|\mathbf{s}\!\cdot\!\mathbf{f}|=2$$, the face diagonal lies in a face that contains the point $$(0,0,0)$$ or $$(1,1,1)$$, so it meets $$\ell_2$$. For intersecting lines the shortest distance is $$0$$, so such diagonals do not influence the maximum.
Therefore any face diagonal having a non-zero distance from $$\ell_2$$ must satisfy $$\mathbf{s}\!\cdot\!\mathbf{f}=0$$ and consequently $$|\mathbf{s}\times\mathbf{f}|=\sqrt6.$$

The vectors orthogonal to $$\mathbf{s}$$ with one zero entry are, up to sign and order,

$$(\;1,-1,0),\;(-1,1,0),\;(1,0,-1),\;(-1,0,1),\;(0,1,-1),\;(0,-1,1).$$

Each of these six vectors represents the two face diagonals on the two parallel faces whose normal is the coordinate axis corresponding to the zero entry. For definiteness consider

$$\ell_1:\;(1,0,1)+s(-1,1,0),\;s\in\mathbb{R}$$

which lies in the face $$z=1$$ and joins the vertices $$(1,0,1)$$ and $$(0,1,1).$$

Step 3 Distance between $$\ell_1$$ and $$\ell_2$$.
Take

$$P=(0,0,0)\in\ell_2,\qquad Q=(1,0,1)\in\ell_1,$$ $$\mathbf{u}=\mathbf{s}=(1,1,1),\qquad\mathbf{v}=\mathbf{f}=(-1,1,0).$$

The standard formula for the shortest distance between two skew lines is

$$d(\ell_1,\ell_2)=\frac{|(Q-P)\cdot(\mathbf{u}\times\mathbf{v})|}{|\mathbf{u}\times\mathbf{v}|}.$$

Compute the cross product: $$\mathbf{u}\times\mathbf{v} =\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\[2pt] 1&1&1\\ -1&1&0 \end{vmatrix} =(-1,-1,2),\qquad|\mathbf{u}\times\mathbf{v}|=\sqrt6.$$

The triple product is $$(Q-P)\cdot(\mathbf{u}\times\mathbf{v}) =(1,0,1)\cdot(-1,-1,2)=(-1)+0+2=1.$$

Hence $$d(\ell_1,\ell_2)=\frac{1}{\sqrt6}.$$

Step 4 Showing this is the maximum.
For every face diagonal that is skew to $$\ell_2$$ the direction vector belongs to the above list, so $$|\mathbf{s}\times\mathbf{f}|=\sqrt6$$. Its two vertices are of the form $$(1,0,1)$$ & $$(0,1,1)$$ (or any permutation/sign-change of these coordinates). A direct substitution (as above) gives

$$|(Q-P)\cdot(\mathbf{s}\times\mathbf{f})|=1$$

for either choice of the end-point $$Q,$$ and therefore

$$d(\ell_1,\ell_2)=\frac{1}{\sqrt6}.$$

Lines with $$|\mathbf{s}\!\cdot\!\mathbf{f}|=2$$ meet $$\ell_2$$, giving distance $$0$$, so no larger value is possible. Since all four body diagonals and all twelve face diagonals are equivalent under cube symmetries, the same calculation applies to every pair in $$S\times F$$.

Thus the maximum distance between a line of $$F$$ and a line of $$S$$ is

$$\boxed{\dfrac{1}{\sqrt6}}$$

Option A is correct.