1. Define the vertices and their properties:
Let the vertices of the dodecagon $S$ be represented by vectors $v_1, v_2, \ldots, v_{12}$ with the origin at the center of the unit circle. Since the dodecagon is inscribed in a unit circle, each vector $v_i$ has a magnitude of 1, i.e., $\|v_i\| = 1$ for all $i$.

2. Calculate the sum of the squares of the distances between all pairs of vertices:
The distance between any two vertices $v_i$ and $v_j$ is given by $\|v_i - v_j\|$. The square of this distance is:
$$\|v_i - v_j\|^2 = (v_i - v_j) \cdot (v_i - v_j) = v_i \cdot v_i + v_j \cdot v_j - 2 v_i \cdot v_j = 1 + 1 - 2 v_i \cdot v_j = 2 - 2 v_i \cdot v_j$$
Therefore, the sum of the squares of the distances between all pairs of vertices is:
$$\sum_{i=1}^{12} \sum_{j=1}^{12} \|v_i - v_j\|^2 = \sum_{i=1}^{12} \sum_{j=1}^{12} (2 - 2 v_i \cdot v_j)$$

3. Simplify the double sum:
$$\sum_{i=1}^{12} \sum_{j=1}^{12} (2 - 2 v_i \cdot v_j) = 2 \sum_{i=1}^{12} \sum_{j=1}^{12} 1 - 2 \sum_{i=1}^{12} \sum_{j=1}^{12} v_i \cdot v_j$$
The first term simplifies to:
$$2 \sum_{i=1}^{12} \sum_{j=1}^{12} 1 = 2 \cdot 12 \cdot 12 = 288$$
The second term involves the dot product of the vectors. Since the vectors are unit vectors and the sum of the dot products of all pairs of vectors in a regular polygon is zero (due to symmetry and orthogonality properties):
$$\sum_{i=1}^{12} \sum_{j=1}^{12} v_i \cdot v_j = 0$$
Therefore, the sum of the squares of the distances is:
$$288 - 0 = 288$$

4. Calculate the quadratic mean (QM) of the distances:
There are $\binom{12}{2} = \frac{12 \cdot 11}{2} = 66$ pairs of vertices. The quadratic mean of the distances is:
$$\text{QM} = \sqrt{\frac{\sum_{i=1}^{12} \sum_{j=1}^{12} \|v_i - v_j\|^2}{66}} = \sqrt{\frac{288}{66}} = \sqrt{\frac{24}{11}}$$

5. Simplify the expression:
$$\sqrt{\frac{24}{11}} = \sqrt{\frac{24}{11}} = \frac{2\sqrt{6}}{\sqrt{11}} = \frac{2\sqrt{66}}{11}$$

The final answer is $\boxed{\frac{2\sqrt{66}}{11}}$.