If none are of $x, y, z$ are zero, then there are $4 \cdot 10^{3}=4000$ ways, since $x y z$ must be positive. Indeed, $(a b c)^{2}=x y z$. So an even number of them are negative, and the ways to choose an even number of 3 variables to be negative is 4 ways. If one of $x, y, z$ is 0 , then one of $a, b, c$ is zero at least. So at least two of $x, y, z$ must be 0 . If all 3 are zero, this gives 1 more solution. If exactly 2 are negative, then this gives $3 \cdot 20$ more solutions. This comes from choosing one of $x, y, z$ to be nonzero, and choosing its value in 20 ways. Our final answer is $4000+60+1=4061$.