12. (2007.02.25) Proof: Take a point $P$ on the plane, and draw line segments $P A=a$, $P B=b$, $P C=c$, such that the angles from ray $P A$ to $P B$, $P B$ to $P C$, and $P C$ to $P A$ are all $120^{\circ}$. Connect $A$, $B$, and $C$ to form $\triangle A B C$, and denote $B C=a^{\prime}$, $C A=b^{\prime}$, $A B=c^{\prime}$. Then the original inequality is equivalent to
$$\left(\sum P A\right)^{2} \cdot \sum \frac{1}{a^{\prime 2}} \geqslant 9$$

Since
$$\left(\sum P A\right)^{2} \geqslant 2 \sum b^{\prime} c^{\prime}-\sum a^{\prime 2}$$
(see Liu Jian, Generalization and Application of Bottema's Inequality, Fujian Middle School Mathematics, 1994, No. 1, 810), it suffices to prove
$$\left(2 \sum b^{\prime} c^{\prime}-\sum a^{\prime 2}\right) \cdot \sum \frac{1}{a^{\prime 2}} \geqslant 9$$

This is a known inequality.