12. Let $P$ be an interior point of $\triangle A B C$, and extend $A P, B P, C P$ to intersect the opposite sides. In the figure, $a, b, c$, $d$ are the lengths of the respective line segments. Given that $a+b+c=43$, $d=3$, find $a b c=$ ?
12. $(441)$
$$
\begin{array}{l} 
\because \frac{S_{\triangle B P C}}{S_{\triangle B A C}}=\frac{d}{d+a}, \frac{S_{\triangle C P A}}{S_{\triangle C B A}}=\frac{d}{d+b}, \\
\frac{S_{\triangle A P B}}{S_{\triangle A C B}}=\frac{d}{d+c},
\end{array}
$$

and $S_{\triangle B P C}+S_{\triangle C P A}+S_{\triangle A P B}=S_{\triangle A B C}$,
$$
\therefore \quad \frac{d}{d+a}+\frac{d}{d+b}+\frac{d}{d+c}=
$$

eliminating the denominators and simplifying, we get
$$
\text { collect } \quad \begin{array}{c}
2 d^{3}+\left(a+b+c=2 d^{2}+a b c=0 .\right. \\
+43 \times 3^{2}=441 .
\end{array}
$$