Solve the following system of equations:

$$
\begin{aligned}
x^{2}+7 y+2 & =2 z+4 \sqrt{7 x-3} \\
y^{2}+7 z+2 & =2 x+4 \sqrt{7 y-3} \\
z^{2}+7 x+2 & =2 y+4 \sqrt{7 z-3}
\end{aligned}
$$
Solution. Let's add the three equations, then rearrange!

$$
\begin{gathered}
x^{2}+7 y+2+y^{2}+7 z+2+z^{2}+7 x+2= \\
=2 z+4 \sqrt{7 x-3}+2 x+4 \sqrt{7 y-3}+2 y+4 \sqrt{7 z-3} \\
\left(x^{2}-2 x+1\right)+\left(y^{2}-2 y+1\right)+\left(z^{2}-2 z+1\right)+(7 x-3-4 \sqrt{7 x-3}+4)+ \\
+(7 y-3-4 \sqrt{7 y-3}+4)+(7 z-3-4 \sqrt{7 z-3}+4)=0 .
\end{gathered}
$$

That is:

$$
\begin{gathered}
(x-1)^{2}+(y-1)^{2}+(z-1)^{2}+ \\
+(\sqrt{7 x-3}-2)^{2}+(\sqrt{7 y-3}-2)^{2}+(\sqrt{7 z-3}-2)^{2}=0
\end{gathered}
$$

Each term on the left side of the equation is non-negative, and their sum can only be 0 if every term is 0.

Therefore, only the triplet $x=y=z=1$ can provide a solution. Substituting this triplet, it turns out that it is indeed a solution.

Thus, the solution to the system of equations is $x=y=z=1$.