Given the parabola $C: y^{2}-2 p x(p>0)$, and a fixed point $A\left(\frac{p}{2}, p\right)$, does there exist a point $T$ in the coordinate plane such that any line $l$ passing through $T$ intersects the parabola $C$ at points $B$ and $C$, and as long as $A$, $B$, and $C$ are not collinear, $\triangle A B C$ can always form a right triangle with $\angle B A C$ as the right angle? If such a point exists, find all such points.
Assume there exists such a point $T$. When $l$ is not perpendicular to the $x$-axis, we can set $l: y=k x+m$, clearly $k \neq 0$. From $\left\{\begin{array}{l}y=k x+m, \\ y^{2}=2 p x,\end{array}\right.$ we get $y^{2}-\frac{2 p}{k} y+\frac{2 p m}{k}=0$.
Let $B\left(x_{1}, y_{1}\right), C\left(x_{2}, y_{2}\right)$, then $\left\{\begin{array}{l}y_{1}+y_{2}=\frac{2 p}{k}, \\ y_{1} y_{2}=\frac{2 p m}{k} .\end{array}\right.$ From $\angle B A C=\frac{\pi}{2}$, we have
$$
\begin{aligned}
0 & =\overrightarrow{A B} \cdot \overrightarrow{A C}=\left(x_{1}-\frac{p}{2}\right)\left(x_{2}-\frac{p}{2}\right)+\left(y_{1}-p\right)\left(y_{2}-p\right) \\
& =\left(\frac{y_{1}^{2}}{2 p}-\frac{p}{2}\right)\left(\frac{y_{2}^{2}}{2 p}-\frac{p}{2}\right)+y_{1} y_{2}-p\left(y_{1}+y_{2}\right)+p^{2} \\
& =\frac{y_{1}^{2} y_{2}^{2}}{4 p^{2}}-\frac{1}{4}\left(y_{1}^{2}+y_{2}^{2}\right)+\frac{1}{4} p^{2}+y_{1} y_{2}-p\left(y_{1}+y_{2}\right)+p^{2} \\
& =\frac{m^{2}}{k^{2}}-\frac{1}{4}\left(\frac{4 p^{2}}{k^{2}}-\frac{4 p m}{k}\right)+\frac{2 p m}{k}-\frac{2 p^{2}}{k}+\frac{5}{4} p^{2}=0,
\end{aligned}
$$

Thus, we can rearrange to get $m^{2}+3 p k m+\left(\frac{5}{4} k^{2}-2 k-1\right) p^{2}=0$, which is $\left[m+\left(\frac{5}{2} k+1\right) p\right]\left[m+\left(\frac{1}{2} k-1\right) p\right]=0$, so $m=-\left(\frac{5}{2} k+1\right) p$ or $m=-\left(\frac{1}{2} k-1\right) p$.

When $m=-\left(\frac{1}{2} k-1\right) p$, $l: y=k x-\left(\frac{1}{2} k-1\right) p=k\left(x-\frac{1}{2} p\right)+p$ always passes through the fixed point $\left(\frac{1}{2} p, p\right)$, which is point $A$;

When $m=-\left(\frac{5}{2} k+1\right) p$, $l: y=k x-\left(\frac{5}{2} k+1\right) p=k\left(x-\frac{5}{2} p\right)-p$ always passes through the fixed point $\left(\frac{5}{2} p,-p\right)$, in this case, there exists a fixed point $T\left(\frac{5}{2} p,-p\right)$ that satisfies the condition.

It is easy to verify that when $l$ passes through the point $T\left(\frac{5}{2} p,-p\right)$ and is perpendicular to the $x$-axis, $\overrightarrow{A B} \cdot \overrightarrow{A C}=0$, i.e., $\angle B A C=\frac{\pi}{2}$.
In summary, there exists a unique fixed point $T\left(\frac{5}{2} p,-p\right)$ that satisfies the condition.