Three, (25 points) Find the equation of the parabola obtained by rotating the parabola $y=x^{2}-2 x-3$ around point $A(3,0)$ by $90^{\circ}$ (Note: $x$ is a quadratic function of $y$).
Three, as shown in Figure 5, $A(3,0)$ is the intersection point of the parabola with the positive half of the $x$-axis. Since
$$
\begin{array}{l}
y=x^{2}-2 x-3 \\
=(x-1)^{2}-4,
\end{array}
$$

the vertex of the original parabola is $P(1,-4)$, and the intersection point of the axis of symmetry with the $x$-axis is $B(1,0)$.
(1) If the original parabola is rotated $90^{\circ}$ clockwise around $A(3,0)$, let the new vertex of the parabola be $P^{\prime}$, and point $B$ rotates to position $B^{\prime}$. Then
$$
B^{\prime} A \perp A B, \triangle P A B \cong \triangle P^{\prime} A B^{\prime} .
$$

Therefore, $\angle B^{\prime}=\angle P B A=90^{\circ}, P^{\prime} B^{\prime}=P B=4$,
$$
B^{\prime} A=B A=O A-O B=3-1=2 \text {. }
$$

It is easy to see that $P^{\prime} B^{\prime} / / x$-axis, and $P^{\prime}$ is in the second quadrant.
Thus, $P^{\prime}(-1,2)$.
According to the vertex form, the equation of the new parabola can be set as
$$
x=a(y-2)^{2}-1 \text {. }
$$

Since the parabola passes through $A(3,0)$, substituting gives $a=1$.
At this time, the equation of the new parabola is
$$
x=(y-2)^{2}-1=y^{2}-4 y+3 \text {. }
$$
(2) If the original parabola is rotated $90^{\circ}$ counterclockwise around $A(3,0)$, using the same method, the equation of the new parabola is
$$
x=-(y+2)^{2}+7=-y^{2}-4 y+3 \text {. }
$$

Combining (1) and (2), the equation of the new parabola is
$$
x=y^{2}-4 y+3 \text { or } x=-y^{2}-4 y+3 \text {. }
$$
(Li Ming, Wuhu County No. 3 Middle School, Anhui, 233300)