3. Find all positive integer triples $(a, b, c)$ such that:
$$
\begin{array}{l}
a^{2}+b^{2}=n[a, b]+n^{2}, \\
b^{2}+c^{2}=n[b, c]+n^{2}, \\
c^{2}+a^{2}=n[c, a]+n^{2}
\end{array}
$$

hold simultaneously for some positive integer $n$.
3. If the greatest common divisor of $a, b, c$ is $d>1$, then substituting into the system of equations gives $d \mid n$. In this case, we can replace $a, b, c, n$ with $\frac{a}{d}, \frac{b}{d}, \frac{c}{d}, \frac{n}{d}$, respectively, and the system of equations still holds. Therefore, we can assume without loss of generality that the greatest common divisor of $a, b, c$ is 1.
From the system of equations, we can solve to get
$$
2 a^{2}=n([a, b]-[b, c]+[c, a]+n).
$$

Thus, $n \mid 2 a^{2}$.
Similarly, $n \mid 2 b^{2}$ and $n \mid 2 c^{2}$.
Since the greatest common divisor of $a, b, c$ is 1, $n$ can only be 1 or 2.
If $n=1$, then
$$
a^{2}+b^{2}=[a, b]+1 \leqslant a b+1.
$$

Hence, $2 a b \leqslant a b+1$.
Similarly, $2 b c \leqslant b c+1$.
Thus, $a=b=c=1$.
If $n=2$, then
$$
a^{2}+b^{2}=2[a, b]+4 \leqslant 2 a b+4.
$$

Hence, $(a-b)^{2} \leqslant 4$, i.e., $|a-b| \leqslant 2$.
Similarly, $|b-c| \leqslant 2$ and $|c-a| \leqslant 2$.
In this case, if two of $a, b, c$ are consecutive, assume without loss of generality that $a=b+1$, then substituting into the first equation gives
$$
a^{2}+(a-1)^{2}=2 a(a-1)+4,
$$

which is a contradiction. Therefore, two of $a, b, c$ cannot be consecutive. Thus, two of them must be equal.
Assume without loss of generality that $a=b$.
Substituting into the first equation gives $a=b=2$, and substituting into the second equation gives $4+c^{2}=2[2, c]+4$, which implies that $c$ is even, but this contradicts the assumption that the greatest common divisor of $a, b, c$ is 1.

In summary, if the greatest common divisor of $a, b, c$ is 1, then the only solution is $a=b=c=1$; if the greatest common divisor of $a, b, c$ is $d$, then $\frac{a}{d}=\frac{b}{d}=\frac{c}{d}=1$.

Therefore, the complete set of solutions to the system of equations is $x=y=z=k$, where $k$ is any positive integer.