One, (20 points) On the Cartesian plane, it is known that the line $y=x+a(-1<a<1)$ intersects the parabola $y=1-x^{2}$ at points $A$ and $B$, and the coordinates of point $C$ are $(1,0)$. Question: For what value of $a$ is the area of $\triangle A B C$ maximized? Find the maximum area of $\triangle A B C$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
From $\left\{\begin{array}{l}y=1-x^{2}, \\ y=x+a\end{array}\right.$ eliminating $y$ gives
$$
x^{2}+x+a-1=0 \text {. }
$$

Thus, $x_{1}+x_{2}=-1, x_{1} x_{2}=a-1$.
Then $|A B|^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$
$$
\begin{array}{l}
=2\left(x_{2}-x_{1}\right)^{2}=2\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right] \\
=2(5-4 a) .
\end{array}
$$

The distance from point $C(1,0)$ to $l_{A B}: x-y+a=0$ is
$$
d=\frac{|1+a|}{\sqrt{2}} \text {. }
$$

Then $S_{\triangle A B C}=\frac{1}{2}|A B| \cdot d$
$$
\begin{array}{l}
=\frac{1}{2} \sqrt{(5-4 a)(1+a)^{2}} \\
=\frac{1}{2} \sqrt{2\left(\frac{5}{2}-2 a\right)(1+a)(1+a)} \\
\leqslant \frac{1}{2} \sqrt{2\left(\frac{\frac{5}{2}-2 a+1+a+1+a}{3}\right)^{3}}=\frac{3 \sqrt{3}}{4} \text {. } \\
\end{array}
$$

Equality holds if and only if $a=\frac{1}{2}$. Therefore, the maximum area of $\triangle A B C$ is $\frac{3 \sqrt{3}}{4}$.