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. $4 \sqrt{2}$
Let $A(a, b), B(-b,-a)$, then we have $\left\{\begin{array}{l}b=3-a^{2}, \\ -a=3-b^{2}\end{array} \Rightarrow a^{4}-6 a^{2}-a+6=0\right.$ $\Rightarrow(a+2)(a-1)\left(a^{2}-a-3\right)=0$.
If $a=-2 \Rightarrow b=-1 \Rightarrow|A B|=3 \sqrt{2}$; If $a=1 \Rightarrow b=2 \Rightarrow|A B|=3 \sqrt{2}$; If $a^{2}-a-3=0 \Rightarrow a=a^{2}-3=-b \Rightarrow A, B$ lie on the line $x+y=0$, which is a contradiction. Therefore, the answer is $C$.