I claim there are no solutions. Assume the contrary. We then obtain, via Vieta's theorem, that \begin{align*} x_1 + x_2 + x_3 &= a^2 \\ x_1x_2 + x_2x_3+x_3x_1 &=b^2 \\ x_1x_2x_3 &=ab-3c. \end{align*}Case 1. $3\mid x_1 x_2x_3$. Assume $3\mid x_1$ without loss. In particular, $3\mid ab$. Then $3\nmid x_2x_3$ as $(x_1,x_2)=(x_1,x_3)=1$. Note that $(x_2,x_3)\equiv(2,2)\pmod{3}$ is the only possibility. But then, $a^2\equiv b^2\equiv 1\pmod{3}$, so $3\nmid ab$. Case 2. $3\nmid x_1x_2x_3$. Note that if $(x_1,x_2,x_3)\equiv (1,1,1)\pmod{3}$ then $3\mid a,b$ but $3\nmid x_1x_2x_3=ab-3c$. Likewise if $(x_1,x_2,x_3)\equiv (2,2,2)\pmod{3}$, we get the same conclusion. So, the remaining cases are exactly two of $x_i$ are congruent to one modulo $3$ and one of them is two modulo $3$, and vice versa. In the latter, that is e.g. when $x_1\equiv 1\pmod{3}$ and $x_2\equiv x_3\equiv 2\pmod{3}$, we get $x_1+x_2+x_3\equiv 2\pmod{3}$, which is not a square. Likewise in the former, e.g. when $x_1\equiv x_2\equiv 1\pmod{3}$ and $x_2\equiv 2\pmod{3}$, we get $b^2\equiv 2\pmod{3}$, which once again is absurd.