The first equation of the given system is fulfilled if and only if the following two conditions are satisfied:
▷ the number $7z$ is integer,
▷ $7z \leq 3x + 5y + 7z < 7z + 1$, i.e. $3x + 5y \in [0, 1)$.

Similarly, the second and third equations are fulfilled if and only if the numbers $7x$ and $7y$ are integers and $3y + 5z$, $3z + 5x \in [0, 1)$.

Now consider any triple of non-negative numbers $(x, y, z)$, which is the solution to the problem. The inequalities $z \geq 0$ and $3z + 5x < 1$ imply $5x < 1$, whence $7x < 7/5 < 2$. This means that non-negative integer $7x$ is equal to one of the numbers $0$ or $1$, i.e., $x \in \{0, 1/7\}$. Similarly, $y, z \in \{0, 1/7\}$.

At this point we have only $2^3 = 8$ triples $(x, y, z)$, which are candidates to solve the problem, so we could test them individually. However, this testing can be avoided by noting that if any two of the numbers $x, y, z$ were equal to $1/7$, one of the expressions $3x+5y$, $3y+5z$, $3z+5x$ would be $8/7$, which is greater than $1$, and that is a contradiction. So, at most one of the numbers $x, y, z$ is equal to $1/7$ and the others are equal to zero. But then each of the three (non-negative) expressions $3x + 5y$, $3y + 5z$, $3z + 5x$ is at most equal to $5/7$, so the conditions stated in the beginning of the solution as equivalence are satisfied and all such triples are solutions.

Conclusion. The problem has exactly $4$ solutions
$$(x, y, z) \in \left\{ (0, 0, 0), \left(\frac{1}{7}, 0, 0\right), \left(0, \frac{1}{7}, 0\right), \left(0, 0, \frac{1}{7}\right) \right\}.$$