Three. (50 points) The sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{1}=3, x_{n+1}=\left[\sqrt{2} x_{n}\right]\left(n \in \mathbf{N}_{+}\right) \text {. }
$$

Find all $n$ such that $x_{n} 、 x_{n+1} 、 x_{n+2}$ form an arithmetic sequence, where $[x]$ denotes the greatest integer not exceeding the real number $x$.

保留源文本的换行和格式，翻译结果如下：

Three. (50 points) The sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{1}=3, x_{n+1}=\left[\sqrt{2} x_{n}\right]\left(n \in \mathbf{N}_{+}\right) \text {. }
$$

Find all $n$ such that $x_{n} 、 x_{n+1} 、 x_{n+2}$ form an arithmetic sequence, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Three, from $x_{n}, x_{n+1}, x_{n+2}$ forming an arithmetic sequence, we know
$$
2\left[\sqrt{2} x_{n}\right]=x_{n}+\left[\sqrt{2}\left[\sqrt{2} x_{n}\right]\right] \text {. }
$$

Also, 
$$
x-1x_{n}+\sqrt{2}\left[\sqrt{2} x_{n}\right]-1 \\
>x_{n}+\sqrt{2}\left(\sqrt{2} x_{n}-1\right)-1 \\
=3 x_{n}-\sqrt{2}-1 .
\end{array}
$$

Combining with equation (1), we get
$$
3 x_{n}-\sqrt{2}-1x_{1} \Rightarrow x_{n} \geqslant 3 . \\
\text { Then } x_{n+1}=\left[\sqrt{2} x_{n}\right]>\sqrt{2} x_{n}-1 \\
=x_{n}+x_{n}(\sqrt{2}-1)-1 \\
\geqslant x_{n}+3(\sqrt{2}-1)-1>x_{n} .
\end{array}
$$

By calculation, we get
$$
\begin{array}{l}
x_{1}=3, x_{2}=4, x_{3}=5, x_{4}=7, \\
x_{5}=9, x_{6}=12, x_{7}=16 .
\end{array}
$$

Therefore, when $n \geqslant 7$, $x_{n}>15$.
Thus, $n \leqslant 6$.
Hence, by conclusion (2), we know that the $n$ satisfying the condition is 1 or 3.