7. Given two moving points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the parabola $x^{2}=4 y$ with $y_{1}+y_{2}=2$ and $y_{1} \neq y_{2}$. If the perpendicular bisector of segment $A B$ intersects the $y$-axis at point $C$, then the maximum value of the area of $\triangle A B C$ is $\qquad$
7. $\frac{16 \sqrt{6}}{9}$.

Let the midpoint of line segment $AB$ be $\left(x_{0}, y_{0}\right)$. Then $y_{0}=1$,
$k_{AB}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1}{4}\left(x_{1}+x_{2}\right)=\frac{x_{0}}{2}$.
Thus, the equation of line $AB$ is $y=\frac{x_{0}}{2}\left(x-x_{0}\right)+1$.
The equation of the perpendicular bisector of line segment $AB$ is
$$
y-1=-\frac{2}{x_{0}}\left(x-x_{0}\right),
$$

which always intersects the $y$-axis at $(0,3)$, i.e., $C(0,3)$.
By solving the system of equations of line $AB$ and the parabola, we get
$$
x^{2}-2 x_{0} x+2 x_{0}^{2}-4=0 \text {. }
$$

Then $\Delta>0$, i.e., $-2<x_{0}<2$, and
$$
x_{1}+x_{2}=2 x_{0}, x_{1} x_{2}=2 x_{0}^{2}-4 \text {. }
$$

Therefore, the area of $\triangle ABC$ is
$$
\begin{array}{l}
=\frac{1}{2} \sqrt{\left(1+k_{AB}^{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right)} \sqrt{x_{0}^{2}+4} \\
=\frac{1}{2} \sqrt{\left(1+\frac{x_{0}^{2}}{4}\right)\left(16-4 x_{0}^{2}\right)\left(x_{0}^{2}+4\right)} \\
=\frac{\sqrt{2}}{4} \sqrt{\left(4+x_{0}^{2}\right)\left(8-2 x_{0}^{2}\right)\left(x_{0}^{2}+4\right)} \\
\leqslant \frac{\sqrt{2}}{4} \sqrt{\left(\frac{4+x_{0}^{2}+8-2 x_{0}^{2}+x_{0}^{2}+4}{3}\right)^{3}} \\
=\frac{16 \sqrt{6}}{9},
\end{array}
$$

with equality holding if and only if $x_{0}= \pm \frac{2 \sqrt{3}}{3}$.
Thus, the maximum area of $\triangle ABC$ is $\frac{16 \sqrt{6}}{9}$.