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.

2(mod 3 ) why?

Dividing through by -1

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.