183. 26 * Find infinitely many integer solutions \(x\), \(y\), \(z\) for the equation \(x^{2}+y^{2}+z^{2}=2 y z+2 z x+2 x y-3\).
By using the equation
$$
\begin{array}{r}
1^{2}+\left(n^{2}-n+1\right)^{2}+\left(n^{2}+n+1\right)^{2}=2\left(n^{2}-n+1\right) \\
+2\left(n^{2}-n+1\right)\left(n^{2}+n+1\right)+2\left(n^{2}+n+1\right)-3,
\end{array}
$$
where $n$ is any integer, we obtain infinitely many integer solutions to the original equation: $x=1, y=n^{2}-n+1, z=n^{2}+n+1$, where $n$ is any integer.

Comment: The original equation can be transformed into $(x-y-z)^{2}=4 y z-3$. Let $t=x-y-z$, so $t^{2}+3=4 y z$. Thus, $t$ is an odd number $2 n-1$, and $n^{2}-n+1=y z$. Taking $y=n^{2}-n+1, z=1$, we get $x=t+y+z=n^{2}+n+1$. This is the solution given above.

Furthermore, since $n^{2}+n+1=y z$, taking $y$ as a divisor of $n^{2}+n+1$, and $z=\frac{n^{2}+n+1}{y}$, we have $x=y+z+2 n-1$, which are all the integer solutions to the equation.