= Solution
Write $\Psi=(|\downarrow\uparrow\rangle+|\uparrow\downarrow\rangle)/\sqrt2$. Each term has cancelling $z$ components, so $S_z\Psi=0$. In the formula for $S^2$, the first three terms contribute $(3/2-1/2)\hbar^2\Psi=\hbar^2\Psi$. The two ladder products exchange the two terms and contribute another $\hbar^2\Psi$. Therefore
$$
\boxed{S^2\Psi=2\hbar^2\Psi,\quad S=1,\quad M_S=0,\quad P_{12}\Psi=+\Psi.}
$$
This <spin> state is symmetric; the accompanying spatial state must be antisymmetric for two <electrons>. A labelled product $|\downarrow_1\uparrow_2\rangle$ alone is not an <eigenstate> of <spin> exchange: its exchanged <vector> is <orthogonal> to it. Thus it cannot supply a definite exchange symmetry when multiplied by a fixed symmetric spatial factor.
This does \b[not] prohibit opposite <spins> in different <spatial orbitals>. For example, with <orthonormal> orbitals $\phi_a,\phi_b$, the <Slater determinant>
$$
\frac1{\sqrt2}\left[\phi_a(\mathbf r_1)\phi_b(\mathbf r_2)|\downarrow_1\uparrow_2\rangle-\phi_b(\mathbf r_1)\phi_a(\mathbf r_2)|\uparrow_1\downarrow_2\rangle\right]
$$
is a valid antisymmetric two-electron state. What fails is treating persistent particle labels as physically distinguishable identities without the required antisymmetrization.
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