= Solution
Writing $A(\mathbf r)=\phi_{1sA}(\mathbf r)$ and $B(\mathbf r)=\phi_{1sB}(\mathbf r)$, the normalized <determinant> is
$$
\boxed{\Psi_{\rm UHF}={1\over\sqrt2}\left[A(\mathbf r_1)\alpha(\omega_1)B(\mathbf r_2)\beta(\omega_2)-B(\mathbf r_1)\beta(\omega_1)A(\mathbf r_2)\alpha(\omega_2)\right].}
$$
Its two <spin orbitals> are orthogonal because $\langle\alpha|\beta\rangle=0$, even if the spatial atomic functions have nonzero overlap. Exchanging both the spatial and spin coordinates of the two <electrons> reverses the sign, as required for fermionic antisymmetry.
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