jmc

Squaring the equation $a+b+c=0,$ we get $$a^2+b^2+c^2+2(ab+ac+bc)=0.$$Hence, $2(ab+ac+bc)=-(a^2+b^2+c^2)\le 0,$ so $$ab+ac+bc\le 0.$$Equality occurs when $a=b=c=0.$

Now, set $c=0,$ so $a+b=0,$ or $b=-a.$ Then $$ab+ac+bc=ab=-a^2$$can take on all nonpositive values. Therefore, the set of all possible values of $ab+ac+bc$ is $\boxed{(-\infty ,0]}.$