What's the probability a $3 \times 3$ circulant matrix with natural coefficients $< n$ is nonsignular?
A circulant matrix $C$ has the form:
$$C = \begin{bmatrix} c_0 & c_1 & c_2 \\ c_2 & c_0 & c_1 \\ c_1 & c_2 & c_0 \end{bmatrix}$$
From Wikipedia's Circulant matrix entry we have that the determinant equals
$$\prod_{k=0}^{n-1}{c_0 + c_1 \omega^{1 k}+ c_2 \omega^{2 k}}$$
where $\omega = e^{2 \pi i / 3}$.
You can assume that each natural number is equally likely to be picked.
Well, you're looking for natural number triples that solve at least one of the equations
$$a+b+c=0 \\ a+be^{2\pi i/3}+ce^{4\pi i/3}=0 \\ a+be^{4\pi i/3} + e^{8\pi i/3}=0$$ where the latter can be rewritten as
$$a+be^{4\pi i/3}+ce^{2\pi i/3}=0.$$
Since you have natural numbers the first one is either unsolvable or has the unique solution $a=b=c=0$ depending on your convention for the natural numbers.
Looking at the second and third ones, they are essentially the same; you simply have that a solution to the second equation can have $b$ and $c$ permuted and yield a solution to third equation instead. That said, if you examine the real and imaginary parts of, say, the second equation, you should arrive at the solution rather quickly.