3. Let non-negative real numbers $a, b, c$ satisfy $a+b+c=3$, and the range of $a^{2}+b^{2}+c^{2}+\lambda a b c$ is the closed interval $[\alpha(\lambda), \beta(\lambda)](\lambda \in \mathbf{R})$. Find the value of $\alpha(\lambda)+\beta(\lambda)$.
Let $a \geqslant b \geqslant c \geqslant 0$. Then $a \geqslant 1 \geqslant c \geqslant 0$. If $a=b=c=1$, then
$$
f(a, b, c)=a^{2}+b^{2}+c^{2}+\lambda a b c=3+\lambda \text {. }
$$

In the following, we assume $a>1>c \geqslant 0$.
(1) If $\lambda \leqslant 2$, fixing $c$, then
$$
f(a, b, c)=(3-c)^{2}+c^{2}+a b(\lambda c-2) \text {; }
$$
(2) If $\lambda>2$, fixing $a$, then
$$
\begin{array}{l}
f(a, b, c)=a^{2}+(3-a)^{2}+b c(\lambda a-2) . \\
\text { Hence } \alpha(\lambda)+\beta(\lambda) \\
=\left\{\begin{array}{ll}
\lambda+12, & \lambda \in\left(-\infty, \frac{3}{2}\right] ; \\
\frac{27}{2}, & \lambda \in\left(\frac{3}{2}, 6\right] ; \\
\lambda+\frac{15}{2}, & \lambda \in(6,+\infty) .
\end{array}\right.
\end{array}
$$