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 through point $P$ a line parallel to the $x$-axis 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 $\triangle A B C$.
(1) The parabola equation is $y^{2}=4 x$, let $A\left(t^{2}, 2 t\right)$,
$$
k_{A C}=\frac{2}{t+1} \Rightarrow A C: y=\frac{2}{t+1}(x-1)+2 \text {. }
$$

Thus $\frac{2}{t+1}(x-1)+2=x+3$
$$
\Rightarrow P\left(\frac{t+3}{1-t}, \frac{6-2 t}{1-t}\right) \Rightarrow B\left(\left(\frac{3-t}{1-t}\right)^{2}, \frac{6-2 t}{1-t}\right) \text {. }
$$

Then $k_{A B}=\frac{4}{\frac{6-2 t}{1-t}+2 t}=\frac{2(1-t)}{3-t^{2}}$.
Thus $A B: y=\frac{2(1-t)}{3-t^{2}}\left(x-t^{2}\right)+2 t \Rightarrow 3 y-2 x=t^{2}(y-2)-t(x-3)$, so the line $A B$ passes through the fixed point $Q(3,2)$.
(2) Since $C Q / / x$ axis, then $S_{A B C}=\left|y_{B}-y_{A}\right|=\left|\frac{6-2 t}{1-t}-2 t\right|$
$=2\left|\frac{3-2 t+t^{2}}{1-t}\right|=2\left|(t-1)+\frac{2}{t-1}\right| \geqslant 4 \sqrt{2}$, equality holds when $|t-1|=\sqrt{2}$.
Therefore, the minimum area of $\triangle A B C$ is $4 \sqrt{2}$.