Three. (25 points) Given the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and its vertex is $C$.
(1) Prove: If $\triangle A B C$ is an isosceles right triangle, then
$$
\Delta=b^{2}-4 a c=4 \text {; }
$$
(2) The quadratic function $y=x^{2}-(2 a+2) x+a^{2}+5 a+3$ intersects the $x$-axis at points $A$ and $B$, and intersects the line $y=3 x-1$ at point $C$. If $\triangle A B C$ is a right triangle, find the value of $a$ (if the quadratic function intersects the linear function at two points, then point $C$ is the one with the smaller $y$-coordinate).
(1) It is only necessary to prove: $|A B|=\frac{\sqrt{\Delta}}{|a|}$, the distance from vertex $C$ to the $x$-axis is $\left|\frac{\Delta}{4 a}\right|$.
Given that $\triangle A B C$ is an isosceles right triangle
$$
\Rightarrow \frac{\sqrt{\Delta}}{|a|}=2\left|\frac{\Delta}{4 a}\right| \Rightarrow \frac{\Delta}{a^{2}}=\frac{\Delta^{2}}{4 a^{2}}
$$
$\Rightarrow \Delta=0$ (discard) or 4.
(2) The quadratic function $y=[x-(a+1)]^{2}+3 a+2$ intersects the $x$-axis at two different points, then
$$
3 a+2<0 \Rightarrow a<-\frac{2}{3}<0 \text {. }
$$

Note that, the vertex of the quadratic function is $(a+1,3 a+2)$, and the vertex must lie on the line $y=3 x-1$, and the point with the minimum $y$-coordinate is the vertex, so $C$ is the vertex of the parabola.
Combining the conclusion from (1), we have
$$
\begin{array}{l}
\Delta=(2 a+2)^{2}-4 \times 1 \times\left(a^{2}+5 a+3\right)=-12 a-8=4 \\
\Rightarrow a=-1 .
\end{array}
$$