= Solution
Use the classes $(E,2C_3,3\sigma_v)$ of the <C3v point group>. For the three hydrogen $1s$ <atomic orbitals>, the <character of a representation> counts orbitals fixed by each <symmetry operation>. Identity fixes three, a nontrivial threefold rotation fixes none, and a vertical mirror fixes one. Therefore
$$
\chi_{\rm H}=(3,0,1).
$$
The <character inner product> gives multiplicities
$$
n_{A_1}=\frac{3+3}{6}=1,\quad
n_{A_2}=\frac{3-3}{6}=0,\quad
n_E=\frac{2\times3}{6}=1.
$$
Thus \b[$\Gamma_{\rm H}=A_1\oplus E$], with the $E$ contribution accounting for two independent <symmetry orbitals>.
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