Three. (20 points) Given that the graph of the linear function $y=a x+b$ passes through the points $A(\sqrt{3}, \sqrt{3}+2), B(-1, \sqrt{3})$, and $C(c, 2-c)$. Find the value of $a-b+c$.

---

The above text has been translated into English, preserving the original text's line breaks and format.
Three, from $\left\{\begin{array}{l}\sqrt{3}+2=\sqrt{3} a+b, \\ \sqrt{3}=-a+b\end{array} \Rightarrow\left\{\begin{array}{l}a=\sqrt{3}-1, \\ b=2 \sqrt{3}-1 .\end{array}\right.\right.$
Therefore, $2-c=a c+b=(\sqrt{3}-1) c+(2 \sqrt{3}-1)$.
Solving for $c$ gives $c=\sqrt{3}-2$.
Thus, $a-b+c=-2$.