Three, (20 points) Given that the two foci of an ellipse are $F_{1}(-3,0), F_{2}(3,0)$, and the line $y=x-1$ is one of its normals (a line passing through the point of tangency and perpendicular to the tangent). Try to find the length of the chord (normal chord) intercepted by this normal on the ellipse.


Three, it is easy to get that the symmetric point of $F_{2}(3, 0)$ about the line $y=x-1$ is $F_{2}^{\prime}(1,2)$, as shown in Figure 1. Let $A$ and $B$ be the points of intersection of the ellipse with the normal line $y=x-1$. Then $A F_{1}$ and $A F_{2}$ are symmetric about the line $A B$. Since $F_{2}^{\prime}$ is the symmetric point of $F_{2}$ about $A B$, $F_{1}$, $F_{2}^{\prime}$, and $A$ are collinear. Therefore, the coordinates of $A$ are $(5,4)$.

It is easy to get $F_{1} A=4 \sqrt{5}, F_{2} A=2 \sqrt{5}$, so the semi-major axis $a$ of the ellipse is $3 \sqrt{5}$. Additionally, the semi-focal distance $c=3$, so the semi-minor axis $b=6$. Therefore,

the equation of the ellipse is
$$
\frac{x^{2}}{45}+\frac{y^{2}}{36}=1 \text {. }
$$

Let the length of the normal chord $A B$ be $t$, then the coordinates of the other endpoint $B$ of the normal chord are $\left(5-\frac{t}{\sqrt{2}}, 4-\frac{t}{\sqrt{2}}\right)$. Substituting into equation (1) gives
$$
4\left(5-\frac{t}{\sqrt{2}}\right)^{2}+5\left(4-\frac{t}{\sqrt{2}}\right)^{2}=180 \text {. }
$$

Solving for the positive root $t=\frac{80}{9} \sqrt{2}$.
Thus, the length of the normal chord is $\frac{80}{9} \sqrt{2}$.