7. Given two lines with a slope of 1, $l_{1}$ and $l_{2}$, passing through the two foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and $l_{1}$ intersects the ellipse at points $A$ and $B$, $l_{2}$ intersects the ellipse at points $C$ and $D$. If quadrilateral $\square A B C D$ satisfies $A C \perp A B$, and the eccentricity of the ellipse is $\frac{\sqrt{u}-\sqrt{v}}{w}\left(u, v, w \in \mathbf{N}_{+}\right.$, and $\sqrt{u} 、 \sqrt{v}$ are in simplest radical form), then $u^{3} v+6 w=$ $\qquad$ .
7.2012.

It is known that $\square A B C D$ is symmetric about the origin $O$. As shown in Figure 1, let $\angle A \dot{F}_{1} F_{2}=\alpha$. Then
. $\tan \alpha=1$
. $\Rightarrow \alpha=45^{\circ}$.
Since $A C \perp A B$, we know
$A C \perp A F_{1}$.
Thus, $\triangle A F_{1} O$ is an

isosceles right triangle. Therefore,
$$
\text { point } A\left(-\frac{c}{2}, \frac{c}{2}\right) \text {. }
$$

Substituting its coordinates into the ellipse equation, we get
$$
\begin{array}{l}
c^{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)=4 \\
\Rightarrow c^{2}\left(2 a^{2}-c^{2}\right)=4 a^{2}\left(a^{2}-c^{2}\right) \\
\Rightarrow e^{2}\left(2-e^{2}\right)=4\left(1-e^{2}\right) \\
\Rightarrow e^{4}-6 e^{2}+4=0 \\
\Rightarrow e^{2}=3 \pm \sqrt{5} .
\end{array}
$$

Since $e<1$, we have
$$
e=\sqrt{3-\sqrt{5}}=\sqrt{\frac{6-2 \sqrt{5}}{2}}=\frac{\sqrt{10}-\sqrt{2}}{2} \text {. }
$$

Therefore, $u^{3} v+6 w=10^{3} \times 2+6 \times 2=2012$.