3・13 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) $[1,9]$.
(China High School Mathematics League, 1986)
[Solution] From $b^{2}+c^{2}+bc \geqslant 0$, and the second equation given in the problem, we get $-6a + 6 \leqslant 0$, which means $a \geqslant 1$. Therefore, we can rule out options (A), (B), and (C).

Multiplying the first equation by 3 and adding the second equation, we get:
$$
\begin{array}{rlrl} 
& & (b-c)^{2} + 3a^{2} - 30a + 27 = 0, \\
& & (b-c)^{2} = -3\left(a^{2} - 10a + 9\right), \\
& & a^{2} - 10a + 9 = & (a-1)(a-9) \leqslant 0 . \\
\text{Thus, } & & 1 \leqslant a \leqslant 9 .
\end{array}
$$

Therefore, the correct choice is (D).