$$
\text { Three. (25 points) In } \triangle A B C \text {, } A B=8+2 \sqrt{6} \text {, }
$$
$B C=7+2 \sqrt{6}, C A=3+2 \sqrt{6}, O, I$ are the circumcenter and incenter of $\triangle A B C$, respectively. Is the length of segment $O I$ a rational number or an irrational number? Make a judgment and explain your reasoning.
Three, OI=2.5 is a rational number.
Let $a, b, c, p, S, R, r$ represent the three sides, the semi-perimeter, the area, the circumradius, and the inradius of $\triangle ABC$, respectively. Thus,
$$
\begin{array}{l}
p=\frac{1}{2}(a+b+c)=9+3 \sqrt{6}, \\
p-a=2+\sqrt{6}, p-b=6+\sqrt{6}, \\
p-c=1+\sqrt{6}.
\end{array}
$$

Then $a b c=(7+2 \sqrt{6})(3+2 \sqrt{6})(8+2 \sqrt{6})$
$$
\begin{array}{l}
=50(12+5 \sqrt{6}), \\
S=\sqrt{p(p-a)(p-b)(p-c)} \\
=\sqrt{(9+3 \sqrt{6})(2+\sqrt{6})(6+\sqrt{6})(1+\sqrt{6})} \\
=\sqrt{3(3+\sqrt{6})(1+\sqrt{6})(2+\sqrt{6})(6+\sqrt{6})} \\
=\sqrt{3(9+4 \sqrt{6})(18+8 \sqrt{6})} \\
=\sqrt{6}(9+4 \sqrt{6})=3(8+3 \sqrt{6}), \\
R=\frac{a b c}{4 S}=\frac{50(12+5 \sqrt{6})}{4 \times 3(8+3 \sqrt{6})} \\
=\frac{5}{6}(3+2 \sqrt{6}), \\
r=\frac{S}{p}=\frac{3(8+3 \sqrt{6})}{9+3 \sqrt{6}}=\frac{6+\sqrt{6}}{3}.
\end{array}
$$

Also, $O I^{2}=R^{2}-2 R r$ $=\left[\frac{5}{6}(3+2 \sqrt{6})\right]^{2}-2 \times \frac{5}{6}(3+2 \sqrt{6}) \times$
$$
\begin{aligned}
& \frac{6+\sqrt{6}}{3} \\
= & \frac{25(33+12 \sqrt{6})}{36}-\frac{5}{9}(30+15 \sqrt{6}) \\
= & \frac{25(33+12 \sqrt{6})-20(30+15 \sqrt{6})}{36} \\
= & \frac{825-600}{36}=\frac{225}{36},
\end{aligned}
$$

Thus, $O I=\sqrt{\frac{225}{36}}=\frac{15}{6}=2.5$.
This is a rational number.