2. In $\triangle A B C$, $|A B|=13,|B C|=14$, $|C A|=15$, and an internal point $P$ satisfies
$$
\overrightarrow{B P} \cdot \overrightarrow{C A}=18, \overrightarrow{C P} \cdot \overrightarrow{B A}=32 \text {. }
$$

Then $\overrightarrow{A P} \cdot \overrightarrow{B C}=$ $\qquad$
2. 14 .

Notice that,
$$
\begin{array}{l} 
\overrightarrow{A P} \cdot \overrightarrow{B C}+\overrightarrow{B P} \cdot \overrightarrow{C A}+\overrightarrow{C P} \cdot \overrightarrow{A B} \\
= \overrightarrow{A P} \cdot(\overrightarrow{B P}-\overrightarrow{C P})+\overrightarrow{B P} \cdot(\overrightarrow{C P}-\overrightarrow{A P})+ \\
\overrightarrow{C P} \cdot(\overrightarrow{A P}-\overrightarrow{B P}) \\
= 0 .
\end{array}
$$

Therefore, $\overrightarrow{A P} \cdot \overrightarrow{B C}=-\overrightarrow{B P} \cdot \overrightarrow{C A}+\overrightarrow{C P} \cdot \overrightarrow{B A}=14$.