Three. (20 points) Given the hyperbola $\Gamma: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. If for any $\triangle A B C$ with vertices all on the hyperbola $\Gamma$, the orthocenter of $\triangle A B C$ is also on the hyperbola $\Gamma$. Explore the condition that the hyperbola $\Gamma$ should satisfy.
Let $A(a \sec \alpha, b \tan \alpha), B(a \sec \beta, b \tan \beta)$, $C(a \sec \gamma, b \tan \gamma)$, and the centroid $H(a \sec \theta, b \tan \theta)$.
Then, $\boldsymbol{A B}=(a(\sec \beta-\sec \alpha), b(\tan \beta-\tan \alpha))$,
$$
\boldsymbol{C H}=(a(\sec \theta-\sec \gamma), b(\tan \theta-\tan \gamma)) \text {. }
$$

By $\boldsymbol{A B} \cdot \boldsymbol{C H}=0$ we get
$$
\begin{array}{l}
a^{2}(\sec \beta-\sec \alpha)\left(\sec \theta-\sec \gamma\right)+ \\
b^{2}(\tan \beta-\tan \alpha)(\tan \theta-\tan \gamma)=0,
\end{array}
$$

which simplifies to
$$
\begin{array}{l}
a^{2}(\cos \alpha-\cos \beta)(\cos \gamma-\cos \theta)+ \\
b^{2} \sin (\beta-\alpha) \cdot \sin (\theta-\gamma)=0 .
\end{array}
$$

From the above equation, we easily obtain
$$
\begin{array}{l}
4 \sin \frac{\alpha-\beta}{2} \cdot \sin \frac{\gamma-\theta}{2}\left(a^{2} \sin \frac{\alpha+\beta}{2} \cdot \sin \frac{\gamma+\theta}{2}+\right. \\
\left.b^{2} \cos \frac{\alpha-\beta}{2} \cdot \cos \frac{\gamma-\theta}{2}\right)=0 .
\end{array}
$$

Given the arbitrariness of $\triangle A B C$, we have
$$
a^{2} \sin \frac{\alpha+\beta}{2} \cdot \sin \frac{\gamma+\theta}{2}+b^{2} \cos \frac{\alpha-\beta}{2} \cdot \cos \frac{\gamma-\theta}{2}=0 \text {, }
$$

which simplifies to
$$
-\frac{1}{2} a^{2}\left(\cos \frac{\alpha+\beta+\gamma+\theta}{2}-\cos \frac{\alpha+\beta-\gamma-\theta}{2}\right)+
$$
$$
\frac{1}{2} b^{2}\left(\cos \frac{\alpha-\beta+\gamma-\theta}{2}+\cos \frac{\alpha-\beta-\gamma+\theta}{2}\right)=0 \text {. }
$$

By $A C \cdot B H=0$, similarly, we get
$$
\begin{array}{l}
-\frac{1}{2} a^{2}\left(\cos \frac{\alpha+\beta+\gamma+\theta}{2}-\cos \frac{\alpha-\beta+\gamma-\theta}{2}\right)+ \\
\frac{1}{2} b^{2}\left(\cos \frac{\alpha+\beta-\gamma-\theta}{2}+\cos \frac{\alpha-\beta-\gamma+\theta}{2}\right)=0 .
\end{array}
$$
Subtracting (2) from (1) gives
$$
\begin{array}{l}
\frac{1}{2} a^{2}\left(\cos \frac{\alpha+\beta-\gamma-\theta}{2}-\cos \frac{\alpha-\beta+\gamma-\theta}{2}\right)+ \\
\frac{1}{2} b^{2}\left(\cos \frac{\alpha-\beta+\gamma-\theta}{2}-\cos \frac{\alpha+\beta-\gamma-\theta}{2}\right)=0
\end{array}
$$

which simplifies to $\frac{1}{2}\left(a^{2}-b^{2}\right)\left(\cos \frac{\alpha+\beta-\gamma-\theta}{2}-\cos \frac{\alpha-\beta+\gamma-\theta}{2}\right)=0$.
Thus, $-\left(a^{2}-b^{2}\right) \sin \frac{\alpha-\theta}{2} \cdot \sin \frac{\beta-\gamma}{2}=0$.
Given the arbitrariness of $\triangle A B C$, we have $a^{2}-b^{2}=0$, i.e., $a=b$. Therefore, the hyperbola $\Gamma$ must be a rectangular hyperbola.