12. Given that $P$ is a point inside $\triangle A B C$ and satisfies $3 P A+4 P B+5 P C=0$. Then, the ratio of the areas of $\triangle P A B$, $\triangle P B C$, and $\triangle P C A$ is $\qquad$
Establish the Cartesian coordinate system as shown in Figure 9. Let
$$
\begin{array}{l}
A(-2,0), \\
B(0,3 x), \\
C(1,2 y), \\
D(3,4) .
\end{array}
$$

Then \( \mid A B \mid \) is in the ratio 12.5:3:4.

Take points \( A^{\prime}, B^{\prime}, C^{\prime} \) on \( P A, P B, P C \) respectively, such that \( P A^{\prime}=3 P A, P B^{\prime}=4 P B, P C^{\prime}=5 P C \). Then \( P A + P B + P C = 0 \), meaning \( P \) is the centroid of \( \triangle A^{\prime} B^{\prime} C^{\prime} \).
Thus, \( S_{\triangle X X^{\prime} B^{\prime}} = S_{\triangle P E C} = S_{\triangle C C A^{\prime}} \).
Also, \( s_{\triangle A B} = \frac{S_{\triangle P A B}}{3 \times 4}, s_{\triangle B B C} = \frac{S_{\triangle P B C}}{4 \times 5} \),
\( s_{\triangle C C A} = \frac{S_{\triangle C K}}{5 \times 3} \),
Therefore, \( S_{\triangle P A B} : S_{\triangle B B C} : S_{\triangle P C A} = 5 : 3 : 4 \).