3. Let real numbers $a, b, c$ satisfy $\left\{\begin{array}{l}a^{2}-b c-8 a+7=0, \\ b^{2}+c^{2}+b c-6 a+6=0,\end{array}\right.$ then the range of values for $a$ is
A. $(-\infty,+\infty)$
B. $(-\infty, 1] \bigcup[9,+\infty)$
C. $(0,7)$
D. $[1,9]$
Adding the two equations, we have $a^{2}-b^{2}-c^{2}-2 b c-2 a+1=0 \Rightarrow (a-1)^{2}=(b+c)^{2}$. Then, $a^{2}-b c-8 a+7=0 \Rightarrow a^{2}-2 a+1-6 a+6=(a-1)^{2}-6(a-1)=b c$
$$
\begin{array}{l}
\leqslant \frac{1}{4}(b+c)^{2}=\frac{1}{4}(a-1)^{2} \Rightarrow \frac{3}{4}(a-1)^{2} \leqslant 6(a-1) \Rightarrow 0 \leqslant a-1 \leqslant 8 \\
\Rightarrow 1 \leqslant a \leqslant 9 \text {. Therefore, the answer is } D .
\end{array}
$$