11. Given the quadratic function $y=a x^{2}+b x+c$ whose graph intersects the $x$-axis at points $A$ and $B$, and whose vertex is $C$.
(1) If $\triangle A B C$ is a right triangle, find the value of $b^{2}-4 a c$.
(2) Consider the quadratic function
$$
y=x^{2}-(2 m+2) x+m^{2}+5 m+3
$$

whose graph intersects the $x$-axis at points $E$ and $F$, and intersects the graph of the linear function $y=3 x-1$ at two points, with the point having the smaller $y$-coordinate denoted as point $G$.
(i) Express the coordinates of point $G$ in terms of $m$;
(ii) If $\triangle E F G$ is a right triangle, find the value of $m$.
3,11. (1) From the given, we have $b^{2}-4 a c>0$.
Let point $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$. Then
$$
|A B|=\left|x_{1}-x_{2}\right|=\frac{\sqrt{b^{2}-4 a c}}{|a|} \text {. }
$$

Note that, the distance from the vertex $C$ to the $x$-axis is $\left|\frac{b^{2}-4 a c}{4 a}\right|$, and $\triangle A B C$ is a right triangle. According to the symmetry of the quadratic function's graph, we have $A C=B C$. Therefore,
$$
\begin{array}{l}
\frac{\sqrt{b^{2}-4 a c}}{|a|}=2\left|\frac{b^{2}-4 a c}{4 a}\right| \\
\Rightarrow \frac{b^{2}-4 a c}{a^{2}}=\frac{\left(b^{2}-4 a c\right)^{2}}{4 a^{2}} \\
\Rightarrow b^{2}-4 a c=4 .
\end{array}
$$
(2) (i) Since the graph of the quadratic function (1) intersects the $x$-axis at two points, we have,
$$
\begin{array}{l}
(2 m+2)^{2}-4\left(m^{2}+5 m+3\right)>0 \\
\Rightarrow 3 m+2<0 \Rightarrow m<-\frac{2}{3} .
\end{array}
$$

Note that,
$$
\begin{array}{l}
y=x^{2}-(2 m+2) x+m^{2}+5 m+3 \\
=[x-(m+1)]^{2}+3 m+2,
\end{array}
$$

its graph opens upwards, and the vertex is $(m+1,3 m+2)$.
It is easy to see that this point must lie on the graph of the function $y=3 x-1$.
Thus, the point $G(m+1,3 m+2)$.
(ii) If $\triangle E F G$ is a right triangle, from the conclusion in (1) we know
$$
\begin{array}{l}
(2 m+2)^{2}-4\left(m^{2}+5 m+3\right) \\
=-12 m-8=4 \\
\Rightarrow m=-1,
\end{array}
$$

which satisfies the condition.
Therefore, $m=-1$.