1. Understanding the Geometry

Let the cube Q have vertices at coordinates $(x_1, x_2, x_3)$ where $x_i \in \{0, 1\}$. This is a unit cube with one vertex at the origin $O(0,0,0)$ and edges along the coordinate axes.

• Set S: The set of 4 lines containing the main diagonals of the cube. These connect opposite vertices. For example, the line passing through $O(0,0,0)$ and $F(1,1,1)$.
• Set F: The set of 12 lines containing the diagonals of the six faces. Each face has two diagonals.

We need to find the maximum value of the shortest distance, $d(\ell_1, \ell_2)$, where $\ell_1 \in F$ and $\ell_2 \in S$.

2. Strategy: Using Symmetry

The cube is a highly symmetric object. We can simplify the problem by fixing one line and then finding its distance to all the lines in the other set. Due to symmetry, the set of distances will be the same regardless of which main diagonal we choose.

Let's choose the main diagonal $\ell_2$ that passes through the origin $O(0,0,0)$ and the vertex $F(1,1,1)$.

• A point on line $\ell_2$ is $\vec{a}_2 = (0,0,0)$.
• The direction vector of $\ell_2$ is $\vec{b}_2 = (1,1,1) - (0,0,0) = (1,1,1)$.

3. Analyzing Distances from $\ell_2$ to lines in F

Now, let's consider a line $\ell_1 \in F$. There are two possibilities for the relationship between $\ell_1$ and $\ell_2$:

Case 1: $\ell_1$ and $\ell_2$ intersect. This occurs if the face diagonal $\ell_1$ shares a vertex with the main diagonal $\ell_2$. The vertices of $\ell_2$ are $O(0,0,0)$ and $F(1,1,1)$.

• There are 3 face diagonals that pass through $O$. For these, the lines intersect at the origin.
• There are 3 face diagonals that pass through $F$. For these, the lines intersect at $F$. For these 6 out of 12 lines in $F$, the shortest distance $d(\ell_1, \ell_2)$ is 0.

Case 2: $\ell_1$ and $\ell_2$ are skew. This occurs for the 6 face diagonals that do not pass through either $O(0,0,0)$ or $F(1,1,1)$. Due to the symmetry of the cube, the shortest distance from $\ell_2$ to each of these 6 lines will be the same. Let's calculate this distance for one of them.

Let's pick the face diagonal $\ell_1$ on the face $z=0$ (the xy-plane) which connects the vertices $A(1,0,0)$ and $C(0,1,0)$.

• A point on line $\ell_1$ is $\vec{a}_1 = (1,0,0)$.
• The direction vector of $\ell_1$ is $\vec{b}_1 = C - A = (0,1,0) - (1,0,0) = (-1,1,0)$.

4. Calculating the Shortest Distance

The formula for the shortest distance between two skew lines with vector equations $\vec{r} = \vec{a}_1 + s\vec{b}_1$ and $\vec{r} = \vec{a}_2 + t\vec{b}_2$ is: $d = \frac{|(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ Let's compute the necessary components:

• $\vec{a}_1 - \vec{a}_2 = (1,0,0) - (0,0,0) = (1,0,0)$.

• $\vec{b}_1 \times \vec{b}_2 = (-1,1,0) \times (1,1,1)$. We can compute this using the determinant: $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(1-0) - \hat{j}(-1-0) + \hat{k}(-1-1) = (1, 1, -2)$ Now, we calculate the numerator and denominator for the distance formula.

• Numerator: $|(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2)| = |(1,0,0) \cdot (1,1,-2)| = |1(1) + 0(1) + 0(-2)| = |1| = 1$

• Denominator: $|\vec{b}_1 \times \vec{b}_2| = |(1,1,-2)| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1+1+4} = \sqrt{6}$

• Shortest Distance: $d(\ell_1, \ell_2) = \frac{1}{\sqrt{6}}$

5. Conclusion

For any pair of lines $(\ell_1, \ell_2)$ with $\ell_1 \in F$ and $\ell_2 \in S$, the shortest distance is either 0 (if they intersect) or $\frac{1}{\sqrt{6}}$ (if they are skew).

The problem asks for the maximum value of $d(\ell_1, \ell_2)$.

Maximum value = $\max \left\{0, \frac{1}{\sqrt{6}}\right\} = \frac{1}{\sqrt{6}}$.

This matches option A.