22. Let positive real numbers $a$, $b$, $c$ satisfy
$$
\left\{\begin{array}{l}
a^{2}+b^{2}=3, \\
a^{2}+c^{2}+a c=4, \\
b^{2}+c^{2}+\sqrt{3} b c=7 .
\end{array}\right.
$$

Find the values of $a$, $b$, and $c$.
22. As shown in Figure 3, from point $O$, draw three line segments $O A$, $O B$, and $O C$ of lengths $a$, $b$, and $c$ respectively, such that $\angle A O B=90^{\circ}$ and $\angle A O C=120^{\circ}$. Then $\angle C O B=150^{\circ}$.
By the cosine rule, we have
$$
A B=\sqrt{a^{2}+b^{2}}=\sqrt{3}, A C=2, B C=\sqrt{7} \text {. }
$$

Thus, $\angle C A B=90^{\circ}$.
Draw $O E \perp A C$ and $O F \perp A B$ from point $O$.
Let $A E=m$ and $O E=n$.
Since $\angle A B O=\angle O A E$, we have
$$
\begin{array}{l}
\triangle B F O \backsim \triangle A E O \\
\Rightarrow \frac{O F}{B F}=\frac{O E}{A E} \Rightarrow \frac{n}{m}=\frac{m}{\sqrt{3}-n} . \\
\text { Also, } \angle A O C=120^{\circ}, \text { then } \\
\tan \angle A O C=-\sqrt{3} \\
\Rightarrow \frac{\frac{m}{n}+\frac{2-m}{n}}{1-\frac{m}{n} \cdot \frac{2-m}{n}}=-\sqrt{3} .
\end{array}
$$

Since $\angle A O C=120^{\circ}$, then

Solving equations (1) and (2) simultaneously, we get $(m, n)=\left(\frac{30}{37}, \frac{12 \sqrt{3}}{37}\right)$.
Thus, $(a, b, c)=\left(\frac{6 \sqrt{37}}{37}, \frac{5 \sqrt{111}}{37}, \frac{8 \sqrt{37}}{37}\right)$.