2. Let $a, b, c \in \mathbf{R}$, and satisfy the system of equations
$$
\left\{\begin{array}{l}
a^{2}+b^{2}+c^{2}-10 a-11=0, \\
a^{2}-b c-4 a-5=0 .
\end{array}\right.
$$

Then the range of values for $a b+b c+c a$ is
$2 \cdot[-40,72]$. From the given, we have
$$
b c=a^{2}-4 a-5, b^{2}+c^{2}=-a^{2}+10 a+11 \text {. }
$$

Then $b+c= \pm \sqrt{(b+c)^{2}}= \pm \sqrt{b^{2}+c^{2}+2 b c}$
$$
= \pm \sqrt{a^{2}+2 a+1}= \pm(a+1) \text {. }
$$

By the relationship between roots and coefficients, $b$ and $c$ are the two real roots of the quadratic equation in $t$, $t^{2} \mp(a+1) t+a^{2}-4 a-5=0$.
By $\Delta=(a+1)^{2}-4\left(a^{2}-4 a-5\right) \geqslant 0$, we solve to get
$-1 \leqslant a \leqslant 7$.
$$
\begin{array}{l}
\text { Let } f(a)=a b+b c+c a=a(b+c)+b c \\
= \pm a(a+1)+a^{2}-4 a-5 \text {, }
\end{array}
$$

Therefore, $f(a)=2(a+1)\left(a-\frac{5}{2}\right)$
or $f(a)=-5(a+1)(-1 \leqslant a \leqslant 7)$.
It is easy to see that when $f(a)=-5(a+1)$, $-40 \leqslant f(a) \leqslant 0$; when $f(a)=2(a+1)\left(a-\frac{5}{2}\right)$, $-\frac{49}{8} \leqslant f(a) \leqslant 72$.