= Solution
In ZZZ order, the only nonzero amplitudes are those of HHH and VVV. Thus
$$
\boxed{P(HHH)=P(VVV)=\frac12,\qquad
P(HHV)=P(HVH)=P(VHH)=P(HVV)=P(VHV)=P(VVH)=0.}
$$
For an equatorial outcome $a=\pm1$, use the X and Y <eigenvectors> in <6/a>. For a product basis containing $k$ Y <quantum measurements>[measurements], the bra acting on VVV contributes $(-i)^kabc/\sqrt8$, while that acting on HHH contributes $1/\sqrt8$. Hence the amplitude of outcome $(a,b,c)$ in the supplied state is
$$
\mathcal A_k(a,b,c)=\frac14[1+(-i)^{k+1}abc].
$$
Each of XXY, XYX and YXX has $k=1$, giving $\mathcal A=(1-abc)/4$. In each of the three respective product bases,
$$
\boxed{|\mathrm{GHZ}\rangle=\frac12\left(
|++-\rangle+|+-+\rangle+|-++\rangle+|---\rangle\right).}
$$
The <eigenvector> in each position belongs to the observable named in that position. All four outcomes with product $abc=-1$ have <probability> $1/4$; the four with $abc=+1$ have <probability> zero. This also specifies all eight <probabilities> in all three bases, and all single-site equatorial marginals are unbiased.
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