Since modulo is "zero" there is no remainder.
$$\frac{(ab-1)(bc-1)(ca-1)}{abc}$$ is not a fraction
$$abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}$$ is not a fraction
So the little terms must sum it up to zero or 1 so,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0$$(let)
$$\frac{ab+bc+ca-1}{abc}=0$$
$$ab+bc+ca=1$$ which certainly can't be true
If,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=1$$
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
But as $$ a,b,c>1$$ $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ has a maxmum value of 1 so there are no solutions to this one either so I arrived at "there are no such a,b,c"
Did I do something wrong?
