3. Let $f=uv\left(u, v \in \mathbf{R}_{+}\right)$. It is known that for all positive triples $(a, b, c)$ satisfying $a+b+c=1$, the inequality
$$
u(ab+bc+ca)+vabc \leqslant \frac{7}{3}
$$

always holds. Then the maximum value of $f$ is . $\qquad$
3. $\frac{441}{4}$.

Given $a+b+c=1, a, b, c \in \mathbf{R}_{+}$, we have $a b+b c+c a \leqslant \frac{(a+b+c)^{2}}{3}=\frac{1}{3}$, and $a b c \leqslant\left(\frac{a+b+c}{3}\right)^{3}=\frac{1}{27}$.
Thus, $u(a b+b c+c a)+v a b c \leqslant \frac{1}{3} u+\frac{1}{27} v$. Let $\frac{1}{3} u+\frac{1}{27} v \leqslant \frac{7}{3}$, which means $9 u+v \leqslant 63$. Then $f=u v=\frac{1}{9} \times 9 u v \leqslant \frac{1}{9}\left(\frac{9 u+v}{2}\right)^{2}$ $\leqslant \frac{1}{9}\left(\frac{63}{2}\right)^{2}=\frac{441}{4}$.
The equality holds when $9 u=v=\frac{63}{2}$, i.e., $u=\frac{7}{2}, v=\frac{63}{2}$.

If there exist $u^{\prime}, v^{\prime} \in \mathbf{R}_{+}, u^{\prime} v^{\prime}>\frac{441}{4}$, such that for all positive numbers $a, b, c$ satisfying $a+b+c=1$, the inequality
$$
u^{\prime}(a b+b c+c a)+v^{\prime} a b c \leqslant \frac{7}{3}
$$
always holds. Taking $a=b=c=\frac{1}{3}$, we get $\frac{1}{3} u^{\prime}+\frac{1}{27} v^{\prime} \leqslant \frac{7}{3}$, which means $9 u^{\prime}+v^{\prime} \leqslant 63$.
However, $9 u^{\prime}+v^{\prime} \geqslant 2 \sqrt{9 u^{\prime} v^{\prime}}>2 \sqrt{9 \times \frac{441}{4}}=63$, a contradiction.
Therefore, $f_{\text {max }}=\frac{441}{4}$.