3. In $\triangle A B C$, it is known that $\overrightarrow{A B} \cdot \overrightarrow{A C}+2 \overrightarrow{B A} \cdot \overrightarrow{B C}=3 \overrightarrow{C A} \cdot \overrightarrow{C B}$, then the minimum value of $\cos C$ is
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{A C}+2 \overrightarrow{B A} \cdot \overrightarrow{B C}=3 \overrightarrow{C A} \cdot \overrightarrow{C B} \Rightarrow b c \cos A+2 c a \cos B=3 a b \cos C \\
\Rightarrow b^{2}+c^{2}-a^{2}+2\left(c^{2}+a^{2}-b^{2}\right)=3\left(a^{2}+b^{2}-c^{2}\right) \Rightarrow a^{2}+2 b^{2}-3 c^{2}=0, \\
\text { then } \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{3 a^{2}+3 b^{2}-\left(a^{2}+2 b^{2}\right)}{6 a b}=\frac{2 a^{2}+b^{2}}{6 a b} \geqslant \frac{\sqrt{2}}{3},
\end{array}
$$

Then $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{3 a^{2}+3 b^{2}-\left(a^{2}+2 b^{2}\right)}{6 a b}=\frac{2 a^{2}+b^{2}}{6 a b} \geqslant \frac{\sqrt{2}}{3}$, equality holds when $b=\sqrt{2} a, c=\frac{\sqrt{15}}{3} a$. Therefore, the minimum value of $\cos C$ is $\frac{\sqrt{2}}{3}$.