Let $Q$ be the cube with the set of vertices $\left\{\right(x_1,x_2,x_3)\in R^3:x_1,x_2,x_3\in \{0,1\left\}\right\}$. 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 correct answer is $\frac{1}{\sqrt{6}}$.

Step 1: Understand the Geometry of the Cube
Let $Q$ be a unit cube with vertices given by $(x_1,x_2,x_3)\in R^3$ where each coordinate is either 0 or 1.

Step 2: Define the Sets of Lines
1. Set $S$ (Main Diagonals):
There are 4 main diagonals passing through opposite vertices of the cube. Consider one such main diagonal ${\ell }_2$ passing through $A(0,0,0)$ and $B(1,1,1)$.
The direction vector of ${\ell }_2$ is:

$\overrightarrow{v_2}=(1,1,1)$

2. Set $F$ (Face Diagonals):
There are 12 face diagonals across the 6 faces of the cube.
Consider a face diagonal ${\ell }_1$ lying on the face $x_3=0$, passing through $C(1,0,0)$ and $D(0,1,0)$.
The direction vector of ${\ell }_1$ is:

$\overrightarrow{v_1}=D-C=(-1,1,0)$

Step 3: Calculate the Shortest Distance Between ${\ell }_1$ and ${\ell }_2$
The formula for the shortest distance $d({\ell }_1,{\ell }_2)$ between two skew lines with direction vectors $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ passing through points $C$ and $A$ respectively is given by:

$d({\ell }_1,{\ell }_2)=\frac{\left|\right(C-A)·(\overrightarrow{v_1}\times \overrightarrow{v_2}\left)\right|}{|\overrightarrow{v_1}\times \overrightarrow{v_2}|}$

First, let's compute the vector cross product $\overrightarrow{v_1}\times \overrightarrow{v_2}$:

$\overrightarrow{v_1}\times \overrightarrow{v_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 1& 0\\ 1& 1& 1\end{array}\right|=\hat{i}(1-0)-\hat{j}(-1-0)+\hat{k}(-1-1)=(1,1,-2)$

Next, calculate the magnitude of the cross product:

$|\overrightarrow{v_1}\times \overrightarrow{v_2}|=\sqrt{1^2+1^2+(-2{)}^2}=\sqrt{1+1+4}=\sqrt{6}$

Now, compute the vector $C-A=(1,0,0)-(0,0,0)=(1,0,0)$.

Compute the scalar dot product:

$(C-A)·(\overrightarrow{v_1}\times \overrightarrow{v_2})=(1,0,0)·(1,1,-2)=1$

Substituting these values back into the shortest distance formula:

$d({\ell }_1,{\ell }_2)=\frac{1}{\sqrt{6}}$

Step 4: Check Other Pairs and Conclusion
For any pair of a face diagonal line ${\ell }_1\in F$ and a main diagonal line ${\ell }_2\in S$, the lines are either intersecting (where distance is 0) or skew lines with a maximum possible shortest distance of $\frac{1}{\sqrt{6}}$.
Thus, the maximum value of $d({\ell }_1,{\ell }_2)$ is $\frac{1}{\sqrt{6}}$.