One, (20 points) Given the quadratic equation in $x$, $a x^{2}+$ $2 a x+2^{2 b}-2^{b+3}+12=0$, cannot have equal real roots, and $\frac{x^{2}}{4}-\sqrt{-a} x+4+c^{2}=0$ has real roots. Find the values of $a, b, c$.
$$
\begin{array}{l}
\left\{\begin{array}{l}
4 a^{2}-4 a\left(2^{3 b}-2^{b+3}+12\right) \leqslant 0, \\
a<0, \\
-a-\left(4+c^{2}\right) \geqslant 0,
\end{array}\right. \\
\text { i.e. }\left\{\begin{array}{l}
a \geqslant 2^{3 b}-2^{b+3}+12, \\
a<0, \\
-4-c^{2} \geqslant a .
\end{array}\right. \\
\therefore-4 \geqslant-4-c^{2} \geqslant a \geqslant 2^{3 b}-2^{b+3}+12 \\
=2^{2 b}-8 \cdot 2^{b}+12 \\
=\left(2^{b}-4\right)^{2}-4 \\
\geqslant-4,
\end{array}
$$

Thus, $a=-4, \left(2^{b}-4\right)^{2}-4=-4, -4-c^{2} \\
=-4$.

Therefore, $a=-4, b$
$$
=2, c=0
$$