12. If positive numbers $a, b, c$ satisfy the system of inequalities
$$
\left\{\begin{array}{l}
\frac{11}{6} c<a+b<2 c, \\
\frac{3}{2} a<b+c<\frac{5}{3} a \\
\frac{5}{2} b<a+c<\frac{11}{4} b .
\end{array}\right.
$$

Determine the order of $a, b, c$.
12. (1) $+c$ gives
$$
\frac{17}{6} c<a+b+c<3 c .
$$
(2) $+a$ gives
$$
\frac{5}{2} a<a+b+c<\frac{8}{3} a \text {. }
$$
(3) $+b$ gives
$$
\frac{7}{2} b<a+b+c<\frac{15}{4} b \text {. }
$$

From (4), (5) we get
$$
\frac{17}{6} c<a+b+c<\frac{8}{3} a,
$$

so $c<\frac{6}{17} \times \frac{8}{3} a=\frac{48}{51} a<1 \cdot a, \quad$ so $c<a$.
Similarly, from (4), (6) we get $b<c$.
Therefore, the order of $a, b, c$ is $b<c<a$.