One. (20 points) Given the parabola $y=a x^{2}+b x+c$ passes through the point $(b,-3)$, and $|a| c+b|c|=0$, and the inequality $a x^{2}+b x+c+3>0$ has no solution. Find all possible values of the triplet $(a, b, c)$.

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From the problem, we know that $(b, -3)$ is the vertex of the parabola, and $a < 0$. Therefore, the parabola is
$$
\begin{array}{l}
y=a(x-b)^{2}-3 \\
\Rightarrow b=-2 a b, c=a b^{2}-3<0 .
\end{array}
$$

From $|a| c+b|c|=0 \Rightarrow|a|=b$.
Thus, $(a, b, c)=\left(-\frac{1}{2}, \frac{1}{2},-\frac{25}{8}\right)$.