Three. (20 points) Let $a_{n}>0\left(n \in \mathbf{N}_{+}\right), a_{1}=5$, when $n \geqslant 2$, $a_{n}+a_{n-1}=\frac{7}{a_{n}-a_{n-1}}+6$. Find the general term formula of the sequence $\left\{a_{n}\right\}$.
The given condition can be transformed into
$$
a_{n}^{2}-a_{n-1}^{2}=6 a_{n}-6 a_{n-1}+7 \text {. }
$$

Completing the square yields $\left(a_{n}-3\right)^{2}-\left(a_{n-1}-3\right)^{2}=7$.
Therefore, $\left(a_{n}-3\right)^{2}=7 n-3$.
Hence, $a_{n}=\sqrt{7 n-3}+3$.