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 $a^{2}+b^{2}+c^{2}-a b-b c-c a$.


$$
\left\{\begin{array} { l } 
{ \sqrt { 3 } + 2 = \sqrt { 3 } a + b , } \\
{ \sqrt { 3 } = - a + b }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=\sqrt{3}-1, \\
b=2 \sqrt{3}-1 .
\end{array}\right.\right.
$$

Solving, we get $c=\sqrt{3}-2$.
Therefore, $a-b=-\sqrt{3}, b-c=\sqrt{3}+1, c-a=-1$.
Thus, $a^{2}+b^{2}+c^{2}-a b-b c-c a$
$$
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=4+\sqrt{3} .
$$