Assume $\{a, b, c\}$ is a triple satisfying the required conditions. For every positive integer $r$, let $S_r = a^r + b^r + c^r$. We denote $S = S_1$.

First, let us show that $S$ divides $S_{11k+1}$ for every non-negative integer $k$. We proceed by induction on $k$. For $k = 0, 1, 2$, the claim is true by our assumptions on $\{a, b, c\}$. Let $k \ge 2$ and assume the result holds for $k-2, k-1$ and $k$; we will see it is true for $k+1$. We have that
$$\begin{aligned} S_{11k+1} \cdot S_{11} &= (a^{11k+1} + b^{11k+1} + c^{11k+1})(a^{11} + b^{11} + c^{11}) \\ &= a^{11(k+1)+1} + b^{11(k+1)+1} + c^{11(k+1)+1} + (ab)^{11}(a^{11(k-1)+1} + b^{11(k-1)+1}) \\ &\quad + (ac)^{11}(a^{11(k-1)+1} + c^{11(k-1)+1}) + (bc)^{11}(b^{11(k-1)+1} + c^{11(k-1)+1}) \\ &= S_{11(k+1)+1} + ((ab)^{11} + (ac)^{11} + (bc)^{11}) S_{11(k-1)+1} - \\ &\quad (abc)^{11}(a^{11(k-2)+1} + b^{11(k-2)+1} + c^{11(k-2)+1}) \\ &= S_{11(k+1)+1} + ((ab)^{11} + (ac)^{11} + (bc)^{11}) S_{11(k-1)+1} - (abc)^{11} S_{11(k-2)+1}. \end{aligned}$$
By the induction assumption, $S$ divides $S_{11k+1}$, $S_{11(k-1)+1}$ and $S_{11(k-2)+1}$; therefore, the above equality implies that $S$ divides $S_{11(k+1)+1}$, as we wanted to prove.

Consider the factorization $x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-xz-yz)$. If $x, y, z$ are integers, it implies that every divisor of $x+y+z$ also divides $x^3+y^3+z^3-3xyz$. By taking $x = a^{11004}$, $y = b^{11004}$ and $z = c^{11004}$, and recalling that $S$ divides $S_{11004}$, we deduce that $S$ divides $S_{33012} - 3(abc)^{11004}$. Since $33012 \equiv 1 \pmod{11}$, we know that $S$ divides $S_{33012}$; therefore, $S$ divides $3(abc)^{11004}$.

Assume $p > 3$ is a prime factor of $S$. Since $p$ divides $3(abc)^{11004}$, then, it divides $a, b$ or $c$. With no loss of generality, assume $p$ divides $a$; then, $p$ divides $b+c$, as it divides $S = a+b+c$. But we also have that $p$ divides $a^{12} + b^{12} + c^{12}$, and looking modulo $p$, we get that $a^{12} + b^{12} + c^{12} \equiv 0^{12} + b^{12} + (-b)^{12} \equiv 2b^{12} \pmod{p}$. It follows that $p$ divides $b$ and, as a consequence, it divides $c$, contradicting the fact that $a, b$ and $c$ are coprime. We conclude that $S$ does not have a prime divisor greater than 3.

Hence, $S = 2^x3^y$ for non-negative integers $x$ and $y$. Finally, we will show that $x, y \le 1$. If $x \ge 2$, we have that $a^{12}+b^{12}+c^{12} \equiv 0 \pmod 4$. As the quadratic residues modulo 4 are 0 and 1, the only possibility is that $a \equiv b \equiv c \equiv 0 \pmod 2$, contradicting the coprimality of $a, b, c$. Similarly, if $y \ge 2$, we have that $a^{12} + b^{12} + c^{12} \equiv 0 \pmod 9$ but, taking into account that for an integer $m$, the possible residues of $m^6$ modulo 9 are 0 and 1, this implies that $a \equiv b \equiv c \equiv 0 \pmod 3$, which is again a contradiction.

Therefore, the possible values of $S = a+b+c$ are 3 and 6, since $a, b, c$ are positive integers and, consequently, the possible triples $\{a, b, c\}$ are $\{1, 1, 1\}$, $\{1, 2, 3\}$, $\{1, 1, 4\}$ and $\{2, 2, 2\}$. It is clear that the first one satisfies the conditions and that the last one is not a solution because $a, b, c$ are not coprime. Now, $\{1, 2, 3\}$ is not a solution either, since $1+2+3=6$ does not divide $1^{12} + 2^{12} + 3^{12}$ (this number has residue 2 modulo 3). To check that $\{1, 1, 4\}$ satisfies the conditions, it suffices to note that $1^r + 1^r + 4^r \equiv 0 \pmod 2$ and $1^r + 1^r + 4^r \equiv 0 \pmod 3$ for every positive integer $r$.

We conclude that the solutions are $\{1, 1, 1\}$ and $\{1, 1, 4\}$.