= Solution
Each <hydrogen> lies on the radial line through its attached <carbon>. Consequently every <molecular symmetry operation> fixes or permutes the <hydrogen> sites exactly as it does the <carbon> sites, and the Cartesian trace is the same. Thus $\Gamma_H=\Gamma_C$ and the complete benzene displacement <group representation> is \b[$\Gamma_{3N}=2\Gamma_C$], of dimension $36$.
Subtract translations $A_{2u}+E_{1u}$ and rotations $A_{2g}+E_{1g}$. The full set of <normal modes> is
$$
\boxed{\begin{aligned}
\Gamma_{\rm vib}={}&2A_{1g}+A_{2g}+2B_{2g}+E_{1g}+4E_{2g}\\
&+A_{2u}+2B_{1u}+2B_{2u}+3E_{1u}+2E_{2u}.
\end{aligned}}
$$
The one-dimensional components contribute $2+1+2+1+2+2=10$ coordinates. The two-dimensional components contribute $2(1+4+3+2)=20$, giving \b[$30=3(12)-6$ vibrational degrees of freedom]. There are twenty distinct symmetry-labelled fundamentals before accidental coincidences, because each doubly degenerate pair has one frequency.
Subtracting translations and rotations directly from the carbon-only result would instead give twelve carbon-skeleton coordinates. That is a six-site mechanical model, not all the <normal modes> of $\mathrm{C_6H_6}$. The <hydrogen> contribution is required for the actual molecule.
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