5. (18th "Hope Cup" Senior High School Competition Problem) Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=0, a_{n+1}=\frac{\sqrt{2} a_{n}+\sqrt{6}}{a_{n}-\sqrt{2}}$ $\left(n \in \mathbf{N}^{*}\right)$, then $a_{2007}=$
A. 0
B. $\sqrt{3}$
C. $-\sqrt{3}$
D. $\sqrt{2}$
5. A Because $a_{4}=0, a_{2}=\frac{\sqrt{6}}{-\sqrt{2}}=-\sqrt{3}, a_{3}=\frac{\sqrt{2}(-\sqrt{3})+\sqrt{6}}{-\sqrt{3}-\sqrt{2}}=0, a_{4}=-\sqrt{3}$, therefore, in the sequence $\left\{a_{n}\right\}$, all odd terms are 0, and all even terms are $-\sqrt{3}$, so $a_{2007}=0$, choose $\mathrm{A}$.