Let real numbers $a, b, c$ satisfy
$$\begin{array}{l}
a^{2}-b c-8 a+7=0 \\
b^{2}+c^{2}+h-6 a+6=0
\end{array}$$

Find the range of real number $a$.
$$\begin{array}{l}
1 b^{2}+c^{2}=-a^{2}+14 a-13, \\
12 b c=2 a^{2}-16 a+14 . \\
\because \quad b^{2}+c^{2} \geqslant 2 b c, \\
\therefore \quad-a^{2}+14 a-13 \geqslant 2 a^{2}-16 a+14,
\end{array}$$

That is, $a^{2}-10 a+9 \leqslant 0$.
Therefore, $1 \leqslant a \leqslant 9$.