10. Given that $a, b, 2$ are the side lengths of a triangle, and $a, b$ are the roots of the equation
$$
\left(3 x^{2}-4 x-1\right)\left(3 x^{2}-4 x-5\right)=12
$$

the perimeter of the triangle can only be ( ).
(A) $\frac{10}{3}$ or $\frac{8}{3}$
(B) $\frac{14}{3}$ or $\frac{10}{3}$
(C) $\frac{16}{3}$ or $\frac{14}{3}$
(D) $\frac{16}{3}$ or $\frac{20}{3}$
10.D.

From $\left(3 x^{2}-4 x\right)^{2}-6\left(3 x^{2}-4 x\right)-7=0$, we get $\left(3 x^{2}-4 x-7\right)\left(3 x^{2}-4 x+1\right)=0$.
Solving, we find $x_{1}=-1, x_{2}=\frac{7}{3}, x_{3}=1, x_{4}=\frac{1}{3}$. If $a \neq b$, from $a+b>2,|a-b|2$, we know $a=b=\frac{7}{3}$.
Thus, the perimeter of the triangle is $\frac{16}{3}$ or $\frac{20}{3}$.