= Solution
Let $a=\phi_{1s}$ and $b=\phi_{2s}$ be orthonormal <atomic orbitals>. The normalized spatial factors are
$$
\Phi_\pm(1,2)=\frac{a(1)b(2)\pm b(1)a(2)}{\sqrt2}.
$$
For the <singlet state>, use the antisymmetric spin factor
$$
\chi_{00}=\frac{\alpha(1)\beta(2)-\beta(1)\alpha(2)}{\sqrt2},\qquad \Psi_S=\Phi_+\chi_{00}.
$$
For the three components of the <triplet state>, use
$$
\chi_{11}=\alpha(1)\alpha(2),\quad
\chi_{10}=\frac{\alpha(1)\beta(2)+\beta(1)\alpha(2)}{\sqrt2},\quad
\chi_{1,-1}=\beta(1)\beta(2),\qquad \Psi_{T,M}=\Phi_-\chi_{1M}.
$$
Each total <wavefunction> obeys the <Pauli principle>. At coincident spatial coordinates,
$$
\Phi_-(\mathbf r,\mathbf r)=\frac{a(\mathbf r)b(\mathbf r)-b(\mathbf r)a(\mathbf r)}{\sqrt2}=0.
$$
Consequently the triplet pair <probability density> vanishes at coincidence, even for its $M=0$ component. This zero is stronger than the trivial statement that any exact point has zero integrated probability in a continuum.
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