3. Find all positive integer triples $\{a, b, c\}$ such that:
(1) $a \leqslant b \leqslant c$ and $(a, b, c)=1$;
(2) $a+b+c$ divides $a^{12}+b^{12}+c^{12}$, $a^{23}+b^{23}+c^{23}$, and $a^{11004}+b^{11004}+c^{11004}$.
3. Let $\{a, b, c\}$ be a set of positive integers that satisfy the given conditions, and let
$$
S_{n}=a^{n}+b^{n}+c^{n}\left(n \in \mathbf{N}_{+}\right), S=S_{1} .
$$

Lemma: For any non-negative integer $k, S$ can divide $S_{11 k+1}$.
Proof by mathematical induction.
For $k=0,1,2$, the conclusion is obviously true according to the problem statement.

When $k \geqslant 2$, assume the proposition holds for $k-2, k-1, k$. Then for $k+1$,
$$
\begin{array}{l} 
S_{11 k+1} S_{11} \\
=\left(a^{11 k+1}+b^{11 k+1}+c^{11 k+1}\right)\left(a^{11}+b^{11}+c^{11}\right) \\
= a^{11(k+1)+1}+b^{11(k+1)+1}+c^{11(k+1)+1}+ \\
(a b)^{11}\left(a^{11(k-1)+1}+b^{11(k-1)+1}\right)+ \\
(a c)^{11}\left(a^{11(k-1)+1}+c^{11(k-1)+1}\right)+ \\
(b c)^{11}\left(b^{11(k-1)+1}+c^{11(k-1)+1}\right) \\
= S_{11(k+1)+1}+ \\
\left((a b)^{11}+(a c)^{11}+(b c)^{11}\right) S_{11(k-1)+1}- \\
(a b c)^{11}\left(a^{11(k-2)+1}+b^{11(k-2)+1}+c^{11(k-2)+1}\right) \\
= S_{11(k+1)+1}+ \\
\left((a b)^{11}+(a c)^{11}+(b c)^{11}\right) S_{11(k-1)+1}- \\
(a b c)^{11} S_{11(k-2)+1},
\end{array}
$$

By the induction hypothesis, $S$ divides $S_{11(k+1)+1}$,
thus the lemma is proved.
Notice,
$$
\begin{array}{l}
x^{3}+y^{3}+z^{3}-3 x y z \\
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-x z-y z\right) \\
\Rightarrow(x+y+z) \mid\left(x^{3}+y^{3}+z^{3}-3 x y z\right) .
\end{array}
$$

Taking $x=a^{11004}, y=b^{11004}, z=c^{11004}$, and combining with
$S$ divides $S_{11004}$, we get
$S \mid\left(S_{33012}-3(a b c)^{11004}\right)$.
Since $33012 \equiv 1(\bmod 11)$, by the lemma,
$S \mid S_{33012}$.
Thus, $S \mid 3(a b c)^{11004}$.
Assume $S$ has a prime factor $p$ greater than 3.
Since $p$ divides $3(a b c)^{11004}$, $p$ must divide one of $a, b, c$.
Without loss of generality, assume $p \mid a$. Then
$$
\begin{array}{l}
p \mid(b+c) \Rightarrow c \equiv-b(\bmod p) . \\
\text { Also, } p \mid\left(a^{12}+b^{12}+c^{12}\right), \text { so } \\
a^{12}+b^{12}+c^{12} \equiv 0^{12}+b^{12}+(-b)^{12} \\
\equiv 2 b^{12}(\bmod p) \\
\Rightarrow p|b \Rightarrow p| c .
\end{array}
$$

This contradicts $(a, b, c)=1$.
Thus, the assumption is false.
Therefore, $S$ has no prime factors greater than 3.
Hence, we can set $S=2^{m} \times 3^{n}(m, n \in \mathbf{N})$.
Next, we prove $m, n \leqslant 1$.
Indeed, if $m \geqslant 2$, then
$$
a^{12}+b^{12}+c^{12} \equiv 0(\bmod 4).
$$

Since the square of any integer modulo 4 can only be 0 or 1, we have
$$
a \equiv b \equiv c \equiv 0(\bmod 2),
$$

which contradicts $(a, b, c)=1$.
Similarly, if $n \geqslant 2$, then
$$
a^{12}+b^{12}+c^{12} \equiv 0(\bmod 9).
$$

Since the sixth power of any positive integer modulo 9 can only be 0 or 1, this implies
$$
a \equiv b \equiv c \equiv 0(\bmod 3),
$$

which also contradicts $(a, b, c)=1$.
Therefore, since $a, b, c$ are positive integers, $S=a+b+c$ can only be 3 or 6, and the triple $\{a, b, c\}$ can only be $\{1,1,1\}, \{1,1,4\}, \{1,2,3\}$, and $\{2,2,2\}$.

Clearly, $\{1,1,1\}$ satisfies the problem's conditions, while $\{2,2,2\}$ does not.
For $\{1,2,3\}$, since $1+2+3=6$, but
$$
1^{12}+2^{12}+3^{12} \equiv 2(\bmod 3),
$$

$6+\left(1^{12}+2^{12}+3^{12}\right)$ does not satisfy the problem's conditions;
For $\{1,1,4\}$, for any positive integer $r$, we have
$1^{r}+1^{r}+4^{r} \equiv 0(\bmod 2)$,
$1^{r}+1^{r}+4^{r} \equiv 0(\bmod 3)$,

so $6 \mid\left(1^{r}+1^{r}+4^{r}\right)$, which satisfies the problem's conditions.
In summary, the only triples that satisfy the conditions are $\{1,1,1\}$

and $\{1,1,4\}$.