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) $(-\infty, 1) \cup[9,+\infty]$
(C) $(0,7)$
(D) $[1,9]$
(D)
3.【Analysis and Solution】From $\left\{\begin{array}{l}a^{2}-b c-8 a+7=0 \\ b^{2}+c^{2}+b c-6 a+6=0,\end{array}\right.$
we get $\left\{\begin{array}{l}b c=a^{2}-8 a+7 \\ b^{2}+c^{2}+b c=6 a-6 .\end{array}\right.$
$$
\begin{array}{l}
\because b^{2}+c^{2}-2 b c=-3\left(a^{2}-10 a+9\right), \text { i.e., }(b+c)^{2}=-3(a-1)(a-9) \geqslant 0, \\
\because 1 \leqslant a \leqslant 9 .
\end{array}
$$