12. Let $a$, $b$, and $c$ be distinct real numbers, and satisfy the following equations:
$$
\begin{array}{l}
b^{2}+c^{2}=2 a^{2}+16 a+14, \\
b c=a^{2}-4 a-5 .
\end{array}
$$

Find the range of values for $a$.
12. From (1) $-2 \times$ (2), we get
$$
(b-c)^{2}=24(a+1)>0 \text {, }
$$

Therefore, $a>-1$.
When $a>-1$, we have
$$
b^{2}+c^{2}=2 a^{2}+16 a+14=2(a+1)(a+7)>0 \text {. }
$$

Furthermore, when $a=b$, from equations (1) and (2) we get
$$
\begin{array}{l}
c^{2}=a^{2}+16 a+14, \\
a c=a^{2}-4 a-5 .
\end{array}
$$

Squaring both sides of equation (1) and combining with equation (3), we get
$$
a^{2}\left(a^{2}+16 a+14\right)=\left(a^{2}-4 a-5\right)^{2} \text {. }
$$

Simplifying, we get $24 a^{3}+8 a^{2}-40 a-25=0$.
$$
\text { Hence }(6 a+5)\left(4 a^{2}-2 a-5\right)=0 \text {. }
$$

Solving, we get $a=-\frac{5}{6}$ or $a=\frac{1 \pm \sqrt{21}}{4}$.
Therefore, the range of $a$ is $a>-1$ and $a \neq-\frac{5}{6}, a \neq$
$$
\frac{1 \pm \sqrt{21}}{4}
$$