15. Given the parabola $y^{2}=2 p x$ passes through the fixed point $C(1,2)$, take any point $A$ on the parabola different from point $C$, the line $A C$ intersects the line $y=x+3$ at point $P$, and a line parallel to the $x$-axis through point $P$ intersects the parabola at point $B$.
(1) Prove: The line $A B$ passes through a fixed point;
(2) Find the minimum value of the area of $\triangle A B C$.
15. As shown in Figure 2.
(1) Given that the parabola $y^{2}=2 p x$ passes through the fixed point $C(1,2)$, the equation of the parabola is $y^{2}=4 x$.
Let point $A\left(\frac{y_{0}^{2}}{4}, y_{0}\right)\left(y_{0} \neq 2\right)$.
Then $l_{A C}: y-2=\frac{y_{0}-2}{\frac{y_{0}^{2}}{4}-1}(x-1)$, which simplifies to
$$
y-2=\frac{4}{y_{0}+2}(x-1),
$$

Solving with $y=x+3$ gives the point
$$
P\left(\frac{-y_{0}-6}{y_{0}-2}, \frac{2 y_{0}-12}{y_{0}-2}\right).
$$

Thus, $B\left(\frac{\left(y_{0}-6\right)^{2}}{\left(y_{0}-2\right)^{2}}, \frac{2 y_{0}-12}{y_{0}-2}\right)$.
When $y_{0}^{2}=12$, $A\left(3, y_{0}\right), B\left(3, \frac{2 y_{0}-12}{y_{0}-2}\right)$, and $l_{A B}$ passes through the fixed point $Q(3,2)$.
$$
\begin{array}{l}
\text { When } y_{0}^{2} \neq 12, \frac{y_{0}^{2}}{4} \neq \frac{\left(y_{0}-6\right)^{2}}{\left(y_{0}-2\right)^{2}}, \\
l_{A B}: y-y_{0}=\frac{y_{0}-\frac{2 y_{0}-12}{y_{0}-2}}{\frac{y_{0}^{2}}{4}-\frac{\left(y_{0}-6\right)^{2}}{\left(y_{0}-2\right)^{2}}}\left(x-\frac{y_{0}^{2}}{4}\right) \\
\Rightarrow y-y_{0}=\frac{y_{0}-2}{y_{0}^{2}-12}\left(4 x-y_{0}^{2}\right).
\end{array}
$$

It is easy to see that $l_{A B}$ also passes through the fixed point $Q(3,2)$.
(2) From (1), we can set $l_{A B}: x-3=m(y-2)$.

Solving with the parabola equation gives
$$
y^{2}-4 m y+4(2 m-3)=0.
$$

Then $y_{1}+y_{2}=4 m, y_{1} y_{2}=4(2 m-3)$,
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2}|C Q|\left|y_{1}-y_{2}\right| \\
=\left|y_{1}-y_{2}\right|=\sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}} \\
=4 \sqrt{(m-1)^{2}+2}.
\end{array}
$$

When $m=1$, the minimum area of $\triangle A B C$ is $4 \sqrt{2}$.