5、 $A, B$ are distinct points on the parabola $y=3-x^{2}$ that are symmetric with respect to the line $x+y=0$, then $|A B|$ equals $($ )
A. 3
B. 4
C. $3 \sqrt{2}$
D. $4 \sqrt{2}$
Solution: C.
[Method 1] Since points $A$ and $B$ are symmetric with respect to the line $x+y=0$, we have $k_{A S}=1$.
Let the equation of line $AB$ be $y=x+b$. Substituting into $y=3-x^{2}$, we get $x^{2}+x+b-3=0$.
From $\Delta=1-4(b-3)>0$, we get $b<\frac{13}{4}$.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, and the midpoint of $AB$ be $M\left(x_{0}, y_{0}\right)$. Then $x_{0}=\frac{x_{1}+x_{2}}{2}=-\frac{1}{2}$. Thus, $y_{0}=x_{0}+b=b-\frac{1}{2}$.
Since point $M\left(-\frac{1}{2}, b-\frac{1}{2}\right)$ lies on the line $x+y=0$, we have $-\frac{1}{2}+b-\frac{1}{2}=0$, which gives $b=1$.
Substituting $b=1$ into (1), we get $x^{2}+x-2=0$. Solving this, we get $x_{1}=-2, x_{1}=1$.
Therefore, $A(-2,-1)$ and $B(1,2)$. Hence, $|A B|=3 \sqrt{2}$.
[Method 2] Since points $A$ and $B$ are symmetric with respect to the line $x+y=0$, we can set $A(a, b)$ and $B(-b,-a)$. Since points $A$ and $B$ lie on the parabola $y=3-x^{2}$, we have
$$
\left\{\begin{array}{l}
b=3-a^{2}, \\
-a=3-(-b)^{2}.
\end{array} \text { Solving this, we get } \left\{\begin{array}{l}
a=-2, \\
b=-1;
\end{array} \text { or } \left\{\begin{array}{l}
a=1, \\
b=2.
\end{array}\right.\right.\right.
$$

Assuming $A(-2,-1)$ and $B(1,2)$, then $|A B|=3 \sqrt{2}$.