6. Given three points $A, B, C$ on a plane satisfying $|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=4,|\overrightarrow{C A}|=5$, then $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}$ is equal to
6. -25 Hint: Since $|\overrightarrow{A B}|^{2}+|\overrightarrow{B C}|^{2}=|\overrightarrow{C A}|^{2}$, then $\angle B=90^{\circ}$. Thus, $\cos \angle A B C=0$,
$$
\begin{array}{l}
\cos \angle B A C=\frac{3}{5}, \cos \angle B C A=\frac{4}{5}, \text { so } \overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}=3 \times 4 \times 0+4 \times 5 \times\left(-\frac{4}{5}\right)+3 \times \\
5 \times\left(-\frac{3}{5}\right)=-25
\end{array}
$$