8. In $\triangle A B C$, $|\overrightarrow{A B}|=2,|\overrightarrow{B C}|=3,|\overrightarrow{C A}|=4$, find the value of $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}$.
8. Since $\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\mathbf{0}$, squaring both sides, we get $\overrightarrow{A B}^{2}+\overrightarrow{B C}^{2}+\overrightarrow{C A}^{2}+2(\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B})=0$. Therefore, $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}=-\frac{1}{2}\left[|\overrightarrow{A B}|^{2}+|\overrightarrow{B C}|^{2}+|\overrightarrow{C A}|^{2}\right]=-\frac{29}{2}$