$\geqslant 3$, side $AB$ is on the $x$-axis. Given that point $Q(0,1)$ is at a distance of 1 from lines $AC$ and $BC$, find the maximum area of $\triangle ABC$.

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Let the coordinates of point $A$ be $(a, 0)$, then the equation of $A C$ is $\frac{x-a}{x_{0}-a}=\frac{y}{y_{0}}$, which simplifies to $-y_{0}(x-a)+y(x_{0}-a)=0$. The distance from point $Q(0,1)$ to $A C$ is 1, so $\frac{\left|a y_{0}+x_{0}-a\right|}{\sqrt{y_{0}^{2}+\left(x_{0}-a\right)^{2}}}=1$, which simplifies to $\left(y_{0}-2\right)a^{2}+2 x_{0} a-y_{0}=0$, (1) The two roots $a_{1} 、 a_{2}$ of (1) are the x-coordinates of $A 、 B$, so
$$
|A B|-\left|a_{1}-a_{2}\right|=\frac{\sqrt{\Delta}}{y_{0}-2}=\frac{2 \sqrt{x_{0}^{2}+y_{0}^{2}-2 y_{0}}}{y_{0}-2} \text {, }
$$

Since $C\left(x_{0}, y_{0}\right)$ satisfies $x_{0}^{2}+y_{0}^{2}-2 y_{0} \leqslant 8, y_{0} \geqslant 3$, we have
$$
S_{\triangle A B C}=\frac{1}{2} \cdot y_{0}|A B|=\frac{y_{0}}{y_{0}-2} \sqrt{x_{0}^{2}+y_{0}^{2}-2 y_{0}} \leqslant \sqrt{8}\left(1+\frac{2}{y_{0}-2}\right) \leqslant 6 \sqrt{2} \text {, }
$$

Thus, the maximum value of $S_{\triangle A B C}$ is $6 \sqrt{2}$ (achieved when $C$ is the point $(\pm \sqrt{5}, 3)$).