125. Solve the system of equations in natural numbers:

$$
\left\{\begin{array}{l}
2 x^{2}+30 y^{2}+3 z^{2}+12 x y+12 y z=308 \\
2 x^{2}+6 y^{2}-3 z^{2}+12 x y-12 y z=92
\end{array}\right.
$$
Solution. By adding the equations of the system and simplifying by 4, we obtain the equation

$$
x^{2}+9 y^{2}+6 x y=100
$$

from which, since $x \in \boldsymbol{N}, y \in \boldsymbol{N}$, we have:

$$
x+3 y=10
$$

After subtracting the second equation from the first, we get:

$$
4 y^{2}+z^{2}+4 y z=36
$$

from which $2 y+z=6$.

150

The solutions to this equation (in natural numbers) are the pairs $(1,4)$ and $(2,2)$. For $y=1$ and $y=2$, from the equation $x+3 y=10$ we get $x=7$ and $x=4$, respectively.

Thus, the solutions to the system are two triples of natural numbers $(7,1,4)$ and $(4,2,2)$.

Answer: $\{(7,1,4),(4,2,2)\}$.