12. Find all triples of distinct positive odd integers $\{a, b, c\}$, such that: $a^{2}+b^{2}+c^{2}=2019$.


Let's assume $a > b > c$, then $a^{2}589 \Rightarrow 24 \leqslant b529=23^{2} \Rightarrow 24 \leqslant b<29$,

and $25^{2}=625 \Rightarrow c^{2} \equiv 3(\bmod 10)$, which is a contradiction, so $a=31$ has no solution;
$$
a=35 \Rightarrow a^{2}=1225 \Rightarrow b^{2}+c^{2}=794 \Rightarrow 397<b^{2}<794 \Rightarrow 19 \leqslant b \leqslant 26,
$$
$b=19 \Rightarrow b^{2}=361 \Rightarrow c^{2} \equiv 3(\bmod 10)$, which is a contradiction; $b=23 \Rightarrow b^{2}=529 \Rightarrow c^{2}=265$, which is a contradiction; $b=25 \Rightarrow b^{2}=625 \Rightarrow c^{2}=169$, which satisfies the condition;
$$
a=37 \Rightarrow a^{2}=1369 \Rightarrow b^{2}+c^{2}=650 \Rightarrow 325<b^{2}<650 \Rightarrow 18 \leqslant b \leqslant 25,
$$
$b=19 \Rightarrow b^{2}=361 \Rightarrow c^{2}=289$, which satisfies the condition; $b=23 \Rightarrow b^{2}=529 \Rightarrow c^{2}=121$, which satisfies the condition;
$b=25 \Rightarrow b^{2}=625 \Rightarrow c^{2}=25$, which satisfies the condition;
$$
a=41 \Rightarrow a^{2}=1681 \Rightarrow b^{2}+c^{2}=338 \Rightarrow 169<b^{2}<338 \Rightarrow 14 \leqslant b \leqslant 18,
$$
$b=17 \Rightarrow b^{2}=289 \Rightarrow c^{2}=49$, which satisfies the condition;
$$
a=43 \Rightarrow a^{2}=1849 \Rightarrow b^{2}+c^{2}=170 \Rightarrow 85<b^{2}<170 \Rightarrow 9 \leqslant b \leqslant 13,
$$
$b=9 \Rightarrow b^{2}=81 \Rightarrow c^{2}=89$, which is a contradiction; $b=11 \Rightarrow b^{2}=121 \Rightarrow c^{2}=49$, which satisfies the condition; $b=13 \Rightarrow b^{2}=169 \Rightarrow c^{2}=1$, which satisfies the condition.
In summary, the solutions that satisfy the condition are $(a, b, c)=(35,25,13),(37,19,17),(37,23,11),(37,25,5),(41,17,7),(43,11,7)$, $(43,13,1)$, a total of 7 sets.