(3) For a point $M(1,1)$ on the parabola $y=x^{2}$ as the right-angle vertex, construct two inscribed right-angled triangles $\triangle M A B$ and $\triangle M C D$ in the parabola, then the coordinates of the intersection point $E$ of line segments $A B$ and $C D$ are
(3) $(-1,2)$ Hint: Let $A\left(x_{1}, x_{1}^{2}\right), B\left(x_{2}, x_{2}^{2}\right)$, then
$$
\begin{array}{l}
k_{M A}=\frac{x_{1}^{2}-1}{x_{1}-1}=x_{1}+1, \\
k_{M B}=\frac{x_{2}^{2}-1}{x_{2}-1}=x_{2}+1, \\
k_{A B}=\frac{x_{1}^{2}-x_{2}^{2}}{x_{1}-x_{2}}=x_{1}+x_{2},
\end{array}
$$

The equation of line $A B$ is
$$
y-x_{1}^{2}=\left(x_{1}+x_{2}\right)\left(x-x_{1}\right),
$$

which simplifies to $y=\left(x_{1}+x_{2}\right) x-x_{1} x_{2}$. Since $M A \perp M B$, then
$$
\left(x_{1}+1\right)\left(x_{2}+1\right)=-1,
$$

which simplifies to
$$
-x_{1} x_{2}=2+\left(x_{1}+x_{2}\right),
$$

Substituting into the equation, we get
$$
y-2=\left(x_{1}+x_{2}\right)(x+1),
$$

Thus, the point $(-1,2)$ lies on line $A B$.
Similarly, if we let $C\left(x_{3}, x_{3}^{2}\right), D\left(x_{4}, x_{4}^{2}\right)$, then the equation of line $C D$ is
$$
y-2=\left(x_{3}+x_{4}\right)(x+1),
$$

which means the point $(-1,2)$ also lies on line $C D$, so the coordinates of intersection point $E$ are $E(-1,2)$.