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 that the line $A B$ passes through a fixed point;
(2) Find the minimum value of the area of $\triangle A B C$.
Solution: (1) Given that the parabola $y^{2}=2 p x$ passes through the fixed point $C(1,2)$, we can obtain the equation of the parabola as $y^{2}=4 x$.
Let the coordinates of point $A$ be $\left(\frac{y_{0}^{2}}{4}, y_{0}\right)\left(y_{0} \neq 2\right)$,
then the equation of line $A C$ is $y-2=\frac{y_{0}-2}{\frac{y_{0}^{2}}{4}-1}(x-1)$,
which simplifies to $y-2=\frac{4}{y_{0}+2}(x-1)$,
and solving it with $y=x+3$ yields the coordinates of point $P$ as $\left(\frac{-y_{0}-6}{y_{0}-2}, \frac{2 y_{0}-12}{y_{0}-2}\right)$.
Therefore, the coordinates of point $B$ are $\left(\frac{\left(y_{0}-6\right)^{2}}{\left(y_{0}-2\right)^{2}}, \frac{2 y_{0}-12}{y_{0}-2}\right)$.
When $y_{0}^{2}=12$, the coordinates of $A$ are $\left(3, y_{0}\right)$, and the coordinates of $B$ are $\left(3, \frac{2 y_{0}-12}{y_{0}-2}\right)$, and line $A B$ passes through the fixed point $Q(3,2)$. When $y_{0}^{2} \neq 12$, $\frac{y_{0}^{2}}{4} \neq \frac{\left(y_{0}-6\right)^{2}}{\left(y_{0}-2\right)^{2}}$, the equation of line $A B$ is $y-y_{0}=\frac{y_{0}-\frac{2 y_{0}-12}{y_{0}-2}}{\frac{y_{0}^{2}}{4}-\frac{\left(y_{0}-6\right)^{2}}{\left(y_{0}-2\right)^{2}}}\left(x-\frac{y_{0}^{2}}{4}\right)$, which simplifies to $y-y_{0}=\frac{\left(y_{0}-2\right)}{y_{0}^{2}-12}\left(4 x-y_{0}^{2}\right)$, (or: $y-y_{0}=\frac{\left(y_{0}-2\right)}{\frac{y_{0}^{2}}{4}-3}\left(x-\frac{y_{0}^{2}}{4}\right)$,) it is easy to see that line $A B$ also passes through the fixed point $Q(3,2)$.
10 points
Method 2: Given that the parabola $y^{2}=2 p x$ passes through the fixed point $C(1,2)$, we can obtain the equation of the parabola as $y^{2}=4 x$.
Let the equation of line $A B$ be $x=m y+a$, and solving it with the parabola equation yields $y^{2}-4 m y-4 a=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then $y_{1}+y_{2}=4 m, y_{1} y_{2}=-4 a$,
the coordinates of point $P$ are $B\left(y_{2}-3, y_{2}\right)$, since $A P$ passes through the fixed point $C$,
we have $\frac{y_{2}-2}{y_{2}-3-1}=\frac{y_{1}-2}{x_{1}-1}$, and $x_{1}=m y_{1}+a$,
so $(m-1) y_{1} y_{2}-(2 m-4) y_{1}+(a+1) y_{2}-2 a-6=0$.
Substituting $y_{1} y_{2}=-4 a, y_{2}=4 m-y_{1}$ into the above equation, we get $(-2 m+3-a) y_{1}+(2 a+4 m-6)=0$. That is, $(-2 m+3-a)\left(y_{1}-2\right)=0$.
Since this equation holds for any $y_{1} \neq 2$, we have $-2 m+3-a=0$, which means $3=2 m+a$, thus the line $x=m y+a$ passes through the fixed point $Q(3,2)$.
... 10 points
(2) From (1), we can set the equation of line $A B$ as $x-3=m(y-2)$,

solving it with the parabola equation yields $y^{2}-4 m y+4(2 m-3)=0$. Then $y_{1}+y_{2}=4 m, y_{1} y_{2}=4(2 m-3)$,
$S_{\triangle A B C}=\frac{1}{2}|C Q| \cdot\left|y_{1}-y_{2}\right|=\left|y_{1}-y_{2}\right|=\sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}}=4 \sqrt{(m-1)^{2}+2}$.
Therefore, when $m=1$, the minimum value of the area of $\triangle A B C$ is $4 \sqrt{2}$. 20 points