10. The function $f(x)=-9 x^{2}-6 a x+2 a-a^{2}$ has a maximum value of -3 on the interval $\left[-\frac{1}{3}, \frac{1}{3}\right]$. Then the value of $a$ is ( ).
(A) $-\frac{3}{2}$
(B) $\sqrt{6}+2$ or $-\sqrt{2}$
(C) $\sqrt{6}+2$ or $2-\sqrt{6}$
(D) $2-\sqrt{6}$ or $-\sqrt{2}$
10. B.

Since $f(x)=-9\left(x+\frac{a}{3}\right)^{2}+2 a$, therefore, when $-\frac{a}{3}>\frac{1}{3}$, i.e., $a>1$, we have
$$
f(x)_{\max }=f\left(-\frac{1}{3}\right)=-1+4 a-a^{2}=-3 \text {. }
$$

Solving this gives $a=\sqrt{6}+2$ (discard $2-\sqrt{6}$).
When $-\frac{1}{3} \leqslant-\frac{a}{3} \leqslant \frac{1}{3}$, i.e., $-1 \leqslant a \leqslant 1$, we have $f(x)_{\max }=f\left(-\frac{a}{3}\right)=2 a=-3$, which has no solution.