2. Let the real number $a$ be such that the inequality $|2 x-a|+|3 x-2 a| \geqslant a^{2}$ holds for any real number $x$. Then the set of $a$ that satisfies the condition is
A. $\left[-\frac{1}{3}, \frac{1}{3}\right]$
B. $\left[-\frac{1}{2}, \frac{1}{2}\right]$
C. $\left[-\frac{1}{4}, \frac{1}{3}\right]$
D. $[-3,3]$
According to the problem, $\min _{x \in \mathbf{R}}\{|2 x-a|+|3 x-2 a|\}=\min \left\{\frac{|a|}{2}, \frac{|a|}{3}\right\}=\frac{|a|}{3}$, thus $a^{2} \leqslant \frac{|a|}{3} \Rightarrow|a| \leqslant \frac{1}{3} \Rightarrow a \in\left[-\frac{1}{3}, \frac{1}{3}\right]$. Therefore, the answer is $A$.