6. Let complex numbers $a, b, c$ satisfy:
$$
|a|=|b|=|c|=2, a+b+c=0 \text {. }
$$

Define $f(z)=|z-a|+|z-b|+|z-c|(z$ being any complex number). Then the minimum value of $f(z)$ is $\qquad$
6. 6 .

Let $a$, $b$, $c$, $z$ correspond to points $A$, $B$, $C$, $P$ in the complex plane.

From the conditions, we know that $\triangle ABC$ is an equilateral triangle inscribed in the circle $\Gamma:|z|=2$, and $P$ is a moving point on the plane. $f(z)$ is the sum of the distances from point $P$ to points $A$, $B$, and $C$.

It is well known that when $P$ is at the Fermat point of $\triangle ABC$, $f(z)$ is minimized. The Fermat point of $\triangle ABC$ is the origin $O$ of the complex plane. Therefore, the minimum value of $f(z)$ is
$$
f(0)=|a|+|b|+|c|=6 .
$$