Solution:
Let $a+b+c=a^{2}+b^{2}+c^{2}=t$. Then $t \geq 0$. On the other hand, the Root mean square - Arithmetic mean inequality implies that
$$\frac{a^{2}+b^{2}+c^{2}}{3} \geq \frac{(a+b+c)^{2}}{9} \Longleftrightarrow 3 t \geq t^{2}$$
Hence $t \in\{0,1,2,3\}$. If $t=0$ or $t=3$, then $a=b=c=0$ or $a=b=c=1$, respectively, and the statement is trivial.

Let $t=1$. Denote by $d$ the product of the denominators of $|a|,|b|$ and $|c|$. Then $x=a d, y=b d$ and $z=c d$ are integers for which $x+y+z=d$ and $x^{2}+y^{2}+z^{2}=d^{2}$. We may assume that $z>0$. Note that
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2} \Longleftrightarrow x y+y z+x z=0 \Longleftrightarrow(x+z)(y+z)=z^{2}$.
It follows that $x+z=r p^{2}$, $y+z=r q^{2}$ and $z=|r| p q$, where $p$ and $q$ are coprime positive integers, and $r$ is a non-zero integer. Since $d=x+y+z= r\left(p^{2}+q^{2}\right)-|r| p q>0$ we get that $r>0$. Then
$$a=\frac{x}{d}=\frac{p(p-q)}{p^{2}+q^{2}-p q}, \quad b=\frac{y}{d}=\frac{q(q-p)}{p^{2}+q^{2}-p q}, \quad c=\frac{z}{d}=\frac{p q}{p^{2}+q^{2}-p q}$$
and thus $a b c=\frac{[p q(p-q)]^{2}}{\left(p q-p^{2}-q^{2}\right)^{3}}$.
It remains to show that $p q(p-q)$ and $p^{2}+q^{2}-p q$ are coprime integers. Suppose that $s$ is a prime divisor of $p q(p-q)$ and $p^{2}+q^{2}-p q$. Let $s \mid p$. Since $s \mid p^{2}+q^{2}-p q$, then $s \mid q$, a contradiction. Analogously, $s \nmid q$. Hence $s \mid p-q$. Then $s \mid(p-q)^{2}-\left(p^{2}+q^{2}-p q\right)=p q$, which is impossible.

The case $t=2$ can be reduced to the case $t=1$ by setting $a_{1}=1-a$, $b_{1}=1-b$ and $c_{1}=1-c$. Indeed, it is easy to check that $a_{1}+b_{1}+c_{1}= a_{1}^{2}+b_{1}^{2}+c_{1}^{2}=1$ and $a_{1} b_{1} c_{1}=-a b c$.