= Solution
Take $p^\mu=(e,0,0,k)$ with $e^2=k^2+m_A^2$. A real orthonormal basis that satisfies the source's stated bilinear normalization is
$$
\boxed{\epsilon_+^\mu=(0,1,0,0),\qquad
\epsilon_-^\mu=(0,0,1,0),\qquad
\epsilon_L^\mu=\frac1{m_A}(k,0,0,e).}
$$
Each obeys $p\cdot\epsilon_i=0$. The two transverse norms are $-1$, while $\epsilon_L^2=(k^2-e^2)/m_A^2=-1$, and all cross products vanish. For <momentum> in the opposite direction, one consistent longitudinal choice is $\epsilon_L(-k)=(k,0,0,-e)/m_A$ for $k>0$; an overall sign of any <polarization> is immaterial.
Here $+$ and $-$ label the two linear transverse basis vectors. If those symbols are intended to denote circular <helicities>, choose $\epsilon_\pm^{\mathrm{hel}}=(\epsilon_x\pm i\epsilon_y)/\sqrt2$ instead. The corresponding orthonormality relation must contain <complex conjugation>, $\epsilon_i^*\cdot\epsilon_j=-\delta_{ij}$. Without that conjugation, a circular <polarization> has $\epsilon_+^{\mathrm{hel}}\cdot\epsilon_+^{\mathrm{hel}}=0$, a concrete counterexample to interpreting the printed bilinear condition as a <helicity> normalization.
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