Ejaifeobuks 10.05.2021 10:58 Taking equation $(mod 3)$ yields $-a^2b^2c^2$ $\equiv 1(mod 3)$ and this we have $a^2b^2c^2$ is $\equiv 2(mod 3)$ which is impossible. Thus there are no solutions.
Albert123 29.10.2021 21:43 In mod $3$ $(3a-bc)(3b-ac)(3c-ab)=1000 \equiv (-bc)(-ac)(-ab) \equiv -a^2b^2c^2 \equiv 1(mod 3)$ $\implies a^2b^2c^2 \equiv 2(mod 3) $ Note that if $x^2 \equiv 1;0(mod 3)$ $\implies a^2b^2c^2 \equiv 1;0(mod 3)$ which is impossible.