jmc

We begin by finding $f(-2)$ and $f(2)$. Since $-2<0$, we have that $f(-2)=(-2{)}^2-2=2$ and since $2\ge 0$, we have that $f(2)=2(2)-20=-16$. Now we can substitute these values back into our equation $f(-2)+f(2)+f(a)=0$ to get $2+(-16)+f(a)=0$, so $f(a)=14$.

Our next step is to find all values of $a$ such that $f(a)=14$. Our first equation $f(a)=a^2-2=14$ yields that $a=\pm 4$, but $a<0$ so $a=-4$ is the only solution. Our second equation $f(a)=2a-20=14$ yields that $a=17$ which is indeed greater than or equal to $0$. Thus, our two possible values of $a$ are $-4$ and $17$ and their positive difference is $17-(-4)=\boxed{21}$.