1. Let real numbers $a$, $b$, $c$ satisfy
$$
\begin{array}{l}
a^{2}-b c-2 a+10=0, \\
b^{2}+b c+c^{2}-12 a-15=0 .
\end{array}
$$

Then the range of values for $a$ is
1. $1 \leqslant a \leqslant 5$.

From $b c=a^{2}-2 a+10$,
$$
(b+c)^{2}=b c+12 a+15 \text {, }
$$

we get $(b+c)^{2}=a^{2}+10 a+25=(a+5)^{2}$.
Then $(b-c)^{2}=(a+5)^{2}-4\left(a^{2}-2 a+10\right) \geqslant 0$,

which means $a^{2}-6 a+5 \leqslant 0$.
Solving this, we get $1 \leqslant a \leqslant 5$.