Three, (15 points) Let the line $l: y=x+c$ intersect the ellipse $\frac{x^{2}}{2}+y^{2}=1$ at two points $A$ and $B$ (which can coincide), and intersect the circle $(x-2)^{2}+(y+2)^{2}=4$ at two points $C$ and $D$ (which can coincide). Find the maximum value of $9|A B|^{2}+|C D|^{2}$.
Three, the chord length formula for the line $l: y=x+c$ intersecting a curve is $\sqrt{2\left(x_{1}+x_{2}\right)^{2}-8 x_{1} x_{2}}$.
Substituting the line equation into the equations of the ellipse and the circle, we get
$$
\begin{array}{l}
\frac{3}{2} x^{2}+2 c x+c^{2}-1=0, \\
2 x^{2}+2 c x+4+4 c+c^{2}=0 .
\end{array}
$$

By Vieta's formulas, we have
$9|A B|^{2}+|C D|^{2}=-18 c^{2}-16 c+32$.
Since the line $l: y=x+c$ intersects the ellipse and the circle, we get the inequality
$$
-\sqrt{3} \leqslant c \leqslant-4+2 \sqrt{2} \text {. }
$$

Since the parabola $y=-18 c^{2}-16 c+32$ has its axis of symmetry at $c=-\frac{8}{18}>-4+2 \sqrt{2}$, it achieves its maximum value $256 \sqrt{2}-336$ at $-4+2 \sqrt{2}$.