3. Given $a, b, c \in \mathbf{R}$, and $a+b+c=3$. Then the minimum value of $3^{a} a+3^{b} b+3^{c} c$ is $\qquad$
3. 9 .

Since the function $y=3^{x}$ is an increasing function on $(-\infty,+\infty)$, by the property of increasing functions, for any $x_{1}$, $x_{2}$, we have $\left(x_{1}-x_{2}\right)\left(3^{x_{1}}-3^{x_{2}}\right) \geqslant 0$, so,
$$
(a-1)\left(3^{a}-3^{1}\right) \geqslant 0,
$$

which means $3^{a} a-3^{\circ} \geqslant 3 a-3$.
Similarly, $3^{b} b-3^{b} \geqslant 3 b-3$, $3^{c} c-3^{c} \geqslant 3 c-3$.
Adding the three inequalities and transforming them, we get
$$
\begin{array}{l}
3^{a} a+3^{b} b+3^{c} c \\
\geqslant 3^{a}+3^{b}+3^{c}+3(a+b+c)-9 \\
=3^{a}+3^{b}+3^{c}+3 \times 3-9=3^{a}+3^{b}+3^{c} . \\
\text { Also, } 3^{a}+3^{b}+3^{c} \geqslant 3 \sqrt[3]{3^{c} \times 3^{b} \times 3^{c}}=3^{2}=9,
\end{array}
$$

with equality holding if and only if $a=b=c=1$.
Therefore, the minimum value of $3^{a} a+3^{b} b+3^{c} c$ is 9.