= Solution
Use $|m_1,m_2\rangle$ for a product of the two individual spin-one <spin eigenstates>, with <magnetic quantum numbers> $m_i=-1,0,1$. Total $S_z=0$ requires $m_1+m_2=0$, leaving precisely
$$
\boxed{|1,-1\rangle,\qquad |0,0\rangle,\qquad |-1,1\rangle.}
$$
Here these are product labels, not labels of the total <angular momentum>. Write their coefficients in this order as $\alpha,\beta,\gamma$. With physical <spin ladder operators> $S_\pm=S_{1\pm}+S_{2\pm}$, each nonzero individual step has <amplitude> $\sqrt2\hbar$, so
$$
S_+|\psi\rangle=\sqrt2\hbar[(\alpha+\beta)|1,0\rangle+(\beta+\gamma)|0,1\rangle],
$$
$$
S_-|\psi\rangle=\sqrt2\hbar[(\alpha+\beta)|0,-1\rangle+(\beta+\gamma)|-1,0\rangle].
$$
Both must vanish for a <singlet state>. The independent <product states> therefore require $\alpha=\gamma=-\beta$. <Wavefunction normalization> gives the <spin-one singlet state>
$$
\boxed{|S=0,M=0\rangle=\frac{|1,-1\rangle-|0,0\rangle+|-1,1\rangle}{\sqrt3},}
$$
up to an overall <quantum phase>. Conversely, $S_z|\psi\rangle=S_+|\psi\rangle=0$ implies $S^2|\psi\rangle=0$ through $S^2=S_-S_++S_z^2+\hbar S_z$, so these conditions really select total <spin> zero. Distinguishability imposes no additional exchange restriction. The ladder hint uses dimensionless <angular momentum> units; restoring its common factor $\hbar$ does not change the coefficient ratios.
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