Three. (20 points) Let the line $y=x+\sqrt{2}$ intersect the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ at points $M$ and $N$, and $O M \perp O N$ ($O$ is the origin). If $M N=\sqrt{6}$, find the equation of the ellipse.
Three, let points $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$.
Substituting $y=x+\sqrt{2}$ into the ellipse equation and eliminating $y$ gives
$$
\begin{array}{l}
\frac{x^{2}}{a^{2}}+\frac{(x+\sqrt{2})^{2}}{b^{2}}=1 \\
\Rightarrow\left(a^{2}+b^{2}\right) x^{2}+2 \sqrt{2} a^{2} x+a^{2}\left(2-b^{2}\right)=0 .
\end{array}
$$

Since points $M, N$ are on the ellipse, i.e., $x_{1}, x_{2}$ are the roots of the equation, we have
$$
x_{1}+x_{2}=\frac{-2 \sqrt{2} a^{2}}{a^{2}+b^{2}}, x_{1} x_{2}=\frac{a^{2}\left(2-b^{2}\right)}{a^{2}+b^{2}} \text {. }
$$

Given $O M \perp O N$
$$
\begin{aligned}
\Rightarrow & x_{1} x_{2}+y_{1} y_{2}=0 \\
\Rightarrow & 0=x_{1} x_{2}+\left(x_{1}+\sqrt{2}\right)\left(x_{2}+\sqrt{2}\right) \\
& =2 x_{1} x_{2}+\sqrt{2}\left(x_{1}+x_{2}\right)+2 .
\end{aligned}
$$

Substituting conclusion (1) into the above equation gives
$$
\begin{array}{l}
\frac{2 a^{2}\left(2-b^{2}\right)}{a^{2}+b^{2}}-\frac{4 a^{2}}{a^{2}+b^{2}}+2=0 \\
\Rightarrow a^{2}+b^{2}=a^{2} b^{2} .
\end{array}
$$

Thus, conclusion (1) becomes
$$
x_{1}+x_{2}=-\frac{2 \sqrt{2}}{b^{2}}, x_{1} x_{2}=\frac{2}{b^{2}}-1 \text {. }
$$

Given $M N=\sqrt{6}$, we have
$$
\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=6 \text {. }
$$

Noting that points $M, N$ lie on the line $y=x+\sqrt{2}$.
$$
\begin{array}{l}
\text { then } x_{1}-x_{2}=y_{1}-y_{2} \\
\Rightarrow\left(x_{1}-x_{2}\right)^{2}=3 \\
\Rightarrow\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=3 .
\end{array}
$$

From conclusion (3), we get
$$
\begin{array}{l}
\frac{8}{b^{4}}-4\left(\frac{2}{b^{2}}-1\right)=3 \\
\Rightarrow b^{4}-8 b^{2}+8=0 .
\end{array}
$$

From equations (2), (4), and $a^{2}>b^{2}$, we get
$$
a^{2}=4+2 \sqrt{2}, b^{2}=4-2 \sqrt{2} \text {. }
$$

Thus, the equation of the ellipse is $\frac{x^{2}}{4+2 \sqrt{2}}+\frac{y^{2}}{4-2 \sqrt{2}}=1$.