 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 isEach obeys $p\cdot {ϵ}_i=0$. The two transverse norms are $-1$, while ${ϵ}_L^2=(k^2-e^2)/m_A^2=-1$, and all cross products vanish. For momentum in the opposite direction, one consistent longitudinal choice is ${ϵ}_L(-k)=(k,0,0,-e)/m_A$ 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 ${ϵ}_{\pm }^{\mathrm{hel}}=({ϵ}_x\pm i{ϵ}_y)/\sqrt{2}$ instead. The corresponding orthonormality relation must contain complex conjugation, ${ϵ}_i^{\ast }\cdot {ϵ}_j=-{\delta }_{ij}$. Without that conjugation, a circular polarization has ${ϵ}_{+}^{\mathrm{hel}}\cdot {ϵ}_{+}^{\mathrm{hel}}=0$, a concrete counterexample to interpreting the printed bilinear condition as a helicity normalization.