3. Let $a, b \in \mathbf{R}$, and $a+b=1$. Then
$$
f(a, b)=3 \sqrt{1+2 a^{2}}+2 \sqrt{40+9 b^{2}}
$$

the minimum value of $f(a, b)$ is $\qquad$
3. $5 \sqrt{11}$.

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
f(a, b) \\
=\sqrt{\left(\frac{81}{11}+\frac{18}{11}\right)\left(1+2 a^{2}\right)}+ \\
\sqrt{\left(\frac{40}{11}+\frac{4}{11}\right)\left(40+9 b^{2}\right)} \\
\geqslant\left(\frac{9}{\sqrt{11}}+\frac{6 a}{\sqrt{11}}\right)+\left(\frac{40}{\sqrt{11}}+\frac{6 b}{\sqrt{11}}\right) \\
=\frac{49}{\sqrt{11}}+\frac{6}{\sqrt{11}}(a+b)=5 \sqrt{11} .
\end{array}
$$

Therefore, the minimum value of $f(a, b)$ is $5 \sqrt{11}$.