Summing the equations gives $a^{2}+b^{2}+c^{2}=0$ and summing $a$ times the first equation and etc. gives $a^{3}+b^{3}+c^{3}=0$. Let $a+b+c=k$. Then $a^{2}+b^{2}+c^{2}=0$ means $a b+b c+c a=k^{2} / 2$, and $a^{3}+b^{3}+c^{3}=0 \Longrightarrow-3 a b c=a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a)=-k^{3} / 2$, so $a b c=k^{3} / 6$. This means $a, b$, and $c$ are roots of the cubic $$x^{3}-k x^{2}+(k^{2} / 2) x-(k^{3} / 6)=0$$ for some $k$. Next, note that $$\begin{aligned} a^{4}+b^{4}+c^{4} & =\sum_{\mathrm{cyc}} a(k a^{2}-(k^{2} / 2) a+(k^{3} / 6)) \\ & =\sum_{\text {cyc }} k(k a^{2}-(k^{2} / 2) a+(k^{3} / 6))-(k^{2} / 2) a^{2}+(k^{3} / 6) a \\ & =\sum_{\text {cyc }}(k^{2} / 2) a^{2}-(k^{3} / 3) a+(k^{4} / 6) \\ & =-k^{4} / 3+k^{4} / 2 \\ & =k^{4} / 6 \end{aligned}$$ After this, there are two ways to extract the values of $k$. - Summing squares of each equation gives $$a^{4}+b^{4}+c^{4}=\sum_{\mathrm{cyc}}(a-b)^{2}=2(a^{2}+b^{2}+c^{2})-2(a b+b c+c a)=-k^{2}$$ so $$\frac{k^{4}}{6}=-k^{2} \Longrightarrow k=0, \pm i \sqrt{6}$$ - Summing $a^{2}$ times the first equation, etc. gives $$a^{4}+b^{4}+c^{4}=\sum_{\mathrm{cyc}} a^{2}(b-c)=-(a-b)(b-c)(c-a)=-a^{2} b^{2} c^{2}=-\frac{k^{6}}{36}$$ so $$\frac{k^{4}}{6}=-\frac{k^{6}}{36} \Longrightarrow k=0, \pm i \sqrt{6}$$ We can achieve $k=0$ with $a=b=c=0$. Letting $a, b$, and $c$ be the roots of $x^{3}-(i \sqrt{6}) x^{2}-3 x+(i \sqrt{6})$ will force one of $a^{2}=b-c$ and all other equalities or $a^{2}-c-b$ and all other equalities to hold, if the latter happens, swap $b$ and $c$. Finally, for these $(a, b, c)$, take $(-a,-c,-b)$ to get $-i \sqrt{6}$. Thus, all of these are achievable.