Solve the following system of equations:

$$
\begin{gathered}
6\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)-49 x y z=0 \\
6 y\left(x^{2}-z^{2}\right)+5 x z=0 \\
2 z\left(x^{2}-y^{2}\right)-9 x y=0
\end{gathered}
$$
In the equations, there is absolutely no constant term, so $(x, y, z) = (0,0,0)$ is a solution to the system of equations. If we expand the multiplications in (2) and (3), we also see that in every term of every equation, at least two of the unknowns appear, so if we set any two unknowns to zero, every term will contain at least one zero factor, thus the equations will be satisfied regardless of the value of the third unknown. Therefore, three solution sets arise, which of course include the previous solution as well.

If one of the three unknowns is set to zero and the other two are not zero, this is impossible, because:

if $x=0$, then from (1) $y^{2} z^{2}=0$, so $y \neq 0, z \neq 0$ is impossible,

if $y=0$, then from (1) $z^{2} x^{2}=0$, so $z \neq 0, x \neq 0$ is impossible,

if $z=0$, then from (1) $x^{2} y^{2}=0$, so $x \neq 0, y \neq 0$ is impossible.

Finally, if neither $x$, nor $y$, nor $z$ is zero, then the solution to the system of equations can be obtained as follows.

Add to (1) the $2y$-multiple of (2) and the $-3z$-multiple of (3), then divide the resulting equation by $6xy$. We get:

$$
3 x y - 2 z = 0, \quad \text{or} \quad z = \frac{3 x y}{2}
$$

Substitute this value of $z$ into (2):

$$
\begin{aligned}
& 6 y \left( x^{2} - \frac{9 x^{2} y^{2}}{4} \right) + 5 x \frac{3 x y}{2} = 0 \quad \mid : x^{2} y \\
& 6 \left( 1 - \frac{9 y^{2}}{4} \right) + \frac{15}{2} = 0 \quad \mid \cdot 4 \\
& 24 - 54 y^{2} + 30 = 0 \\
& y^{2} = 1 \quad y = \pm 1
\end{aligned}
$$

Substituting the value of $z$ from (4) into (3), simplifying by $3xy$, and substituting $y$, we get

$$
x^{2} = 4 \quad x = \pm 2
$$

Finally, for the value of $z$, we get: $z = \frac{3 x y}{2} = \frac{3 (\pm 2) (\pm 1)}{2} = \pm 3$.

So far, nothing has been said about the signs of the roots. The sign determination can be served by the conditional equation (4). According to this, $\frac{z}{xy} = \frac{3}{2}$, which is positive, so either all three signs are positive, or two are negative and one is positive.

In several papers, every sign and justification appears without any explanation, for example, $x=2$, it cannot be known, these are only the results of calculations not included here, or values pulled out of thin air. The discussion of the signs of the roots of the $(\pm 2, \pm 1, \pm 3)$ system of equations is missing in one of the papers.