= Solution
The <trihydrogen cation> is a nonlinear three-nucleus molecule. Removing three translations and three rigid rotations leaves $3(3)-6=3$ vibrational degrees of freedom. One <normal mode> is the symmetric in-plane breathing motion: every nucleus moves radially toward or away from the centre. The remaining two <normal modes> form a degenerate in-plane distortion pair, stretching one direction while compressing another.
For an explicit sketch, let $\mathbf r_j$ be the equilibrium vertices relative to their centre. Displacements proportional to
$$
\mathbf u_j^{(1)}=\mathbf r_j,\qquad
\mathbf u_j^{(2)}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\mathbf r_j,\qquad
\mathbf u_j^{(3)}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\mathbf r_j
$$
give the breathing mode and a basis of the degenerate pair. Linear combinations within the latter pair are equally valid. Their actual frequencies require molecular force constants, which are not supplied. An apparent out-of-plane motion of three point nuclei is a translation or infinitesimal rigid rotation, rather than an additional independent vibration.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2010/paper-2a-triangular-modes.png]
{title=Breathing and degenerate distortion modes of an equilateral molecule}
{description=Original displacement vectors in a simple symmetric bond-spring illustration. The degeneracy is fixed by symmetry; no measured molecular frequencies are asserted.}
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