(12) $O, A, B$ are three non-collinear points on a plane, vector $\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}$, let $P$ be any point on the perpendicular bisector of line segment $A B$, vector $\overrightarrow{O P}=\vec{p}$. If $|\vec{a}|=5,|\vec{b}|=3$, then the value of $\vec{p} \cdot(\vec{a}-\vec{b})$ is $\qquad$.
(12) As shown in the figure, $Q P$ is the perpendicular bisector of line segment $A B$. Since $\overrightarrow{O P}=\overrightarrow{O Q}+\overrightarrow{Q P}, \overrightarrow{O Q}=$ $\frac{1}{2}(\vec{a}+\vec{b}), \overrightarrow{Q P} \perp \overrightarrow{B A}$, therefore
$$
\begin{aligned}
\vec{p} \cdot(\vec{a}-\vec{b}) & =(\overrightarrow{O Q}+\overrightarrow{Q P}) \cdot \overrightarrow{B A} \\
& =\overrightarrow{O Q} \cdot \overrightarrow{B A}+\overrightarrow{Q P} \cdot \overrightarrow{B A} \\
& =\frac{1}{2}(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b}) \\
& =\frac{1}{2}\left(|\vec{a}|^{2}-|\vec{b}|^{2}\right) \\
& =8 .
\end{aligned}
$$