Problem 738

Three, the number of vectors generated is $\mathrm{C}_{7}^{2}=21$, then we can take $\mathrm{C}_{21}^{2}=210$ pairs of vectors. Let the vectors taken be $a, b$. Since $\xi=|a \cdot b|=|a| \cdot|b| \cdot|\cos \langle a, b\rangle|$, therefore, the direction of the vectors can be ignored. Without loss of generality, let the angle between the two vectors be the angle between their respective lines (with a range of $\left[0^{\circ}, 90^{\circ}\right]$), then the angle between any two vectors belongs to the set $\left\{0^{\circ}, 30^{\circ}, 60^{\circ}, 90^{\circ}\right\}$, and the magnitude of each vector belongs to the set $\{1, \sqrt{3}, 2\}$, where the number of vectors with a magnitude of 1 is 12, the number of vectors with a magnitude of $\sqrt{3}$ is 6, and the number of vectors with a magnitude of 2 is 3.

If $|a|=|b|=2$, then the angle between them must be $60^{\circ}$, $|a \cdot b|=2$, and the probability is $\frac{1}{2} \times \frac{3 \times 2}{210}=\frac{1}{70}$.

If $|\boldsymbol{a}|=|\boldsymbol{b}|=\sqrt{3}$, then the angle between them can be $0^{\circ}$ or $60^{\circ}$. When the angle is $0^{\circ}$, $|a \cdot b|=3$, and the probability is $\frac{1}{2} \times \frac{6 \times 1}{210}=\frac{1}{70}$; when the angle is $60^{\circ}$, $|a \cdot b|=\frac{3}{2}$, and the probability is $\frac{1}{2} \times \frac{4 \times 6}{210}=\frac{2}{35}$.

If $|\boldsymbol{a}|=|\boldsymbol{b}|=1$, then the angle between them can be $0^{\circ}$ or $60^{\circ}$. It is easy to know that their probabilities are
$$\frac{1}{2} \times \frac{3 \times 12}{210}=\frac{3}{35}, \frac{1}{2} \times \frac{8 \times 12}{210}=\frac{8}{35} \text {. }$$

If $|a|=2,|\boldsymbol{b}|=\sqrt{3}$, then the angle between them can be $30^{\circ}$ or $90^{\circ}$. It is easy to know that their probabilities are $\frac{3 \times 4}{210}=\frac{2}{35}, \frac{3 \times 2}{210}=\frac{1}{35}$.

If $|a|=2,|b|=1$, then the angle between them can be $0^{\circ}$ or $60^{\circ}$. It is easy to know that their probabilities are $\frac{3 \times 4}{210}=\frac{2}{35}, \frac{3 \times 8}{210}=\frac{4}{35}$.

If $|\boldsymbol{a}|=\sqrt{3},|\boldsymbol{b}|=1$, then the angle between them can be $30^{\circ}$ or $90^{\circ}$. It is easy to know that their probabilities are $\frac{6 \times 8}{210}=\frac{8}{35}, \frac{6 \times 4}{210}=\frac{4}{35}$.

Thus, the probability distribution of $\xi$ is shown in Table 1: Therefore, the mathematical expectation of $\xi$ is
$$\begin{array}{l} E \xi=0 \times \frac{1}{7}+\frac{1}{2} \times \frac{8}{35}+1 \times \frac{1}{5}+ \\ \quad \frac{3}{2} \times \frac{2}{7}+2 \times \frac{1}{14}+3 \times \frac{1}{14} \\ =\frac{11}{10} . \end{array}$$