= Solution
Translate the <generalized coordinates> so that the <equilibrium> is at zero. Its potential gradient vanishes, and its quadratic <potential energy> is $q^T\mathsf Vq/2$, where $\mathsf V$ is the symmetric <Hessian matrix>. The <Euler-Lagrange equations> for small motion are
$$
\mathsf T\ddot q+\mathsf Vq=0.
$$
A <normal mode> has fixed component ratios, $q(t)=v\cos(\omega t+\phi)$, with
$$
\mathsf Vv=\omega^2\mathsf Tv.
$$
The <angular frequency> $\omega$ is its <eigenfrequency>. A zero mode instead admits a constant or linear time factor. For modes $v_\alpha,v_\beta$, symmetry gives
$$
v_\alpha^T\mathsf Vv_\beta=\omega_\beta^2v_\alpha^T\mathsf Tv_\beta
=\omega_\alpha^2v_\alpha^T\mathsf Tv_\beta.
$$
Hence <mass-matrix orthogonality of normal modes> is
$$
\boxed{v_\alpha^T\mathsf Tv_\beta=0\quad\text{if }\omega_\alpha^2\ne\omega_\beta^2.}
$$
Within a degenerate eigenspace, a basis can be chosen orthogonal in this positive <weighted inner product>.
For three masses connected pairwise, the exact energies are
$$
T=\frac m2\sum_{i=1}^3|\dot x_i|^2,\qquad
V=\frac k2\sum_{i<j}(|x_i-x_j|-\sqrt3)^2.
$$
A fixed common <translation> $x_i\mapsto x_i+a$ preserves all differences and velocities. A fixed rigid <rotation> $x_i\mapsto Qx_i$ preserves their <Euclidean norms>. Thus both energies have these symmetries. Here the transformation is fixed in time; a moving reference frame would change the kinetic energy. Every equilateral configuration of side $\sqrt3$ has zero spring extensions, hence zero force and minimum potential, proving that it is an <equilibrium>.
For the specified displacement coordinates, write
$$
\begin{aligned}
d_{23}&=\sqrt{(\sqrt3+q_3-q_5)^2+(q_4-q_6)^2},\\
d_{13}&=\sqrt{(\sqrt3/2+q_1-q_5)^2+(3/2+q_2-q_6)^2},\\
d_{12}&=\sqrt{(-\sqrt3/2+q_1-q_3)^2+(3/2+q_2-q_4)^2}.
\end{aligned}
$$
The exact potential is $V=(k/2)\sum_{ij=12,13,23}(d_{ij}-\sqrt3)^2$. For an equilibrium edge vector $d$ and small perturbation $\delta d$,
$$
|d+\delta d|-|d|=\frac d{|d|}\cdot\delta d+O(|\delta d|^2).
$$
Thus only each first-order extension is needed for the quadratic potential. These extensions, with an immaterial sign choice for the horizontal edge, are
$$
s_1=q_5-q_3,\qquad
s_2=\frac12(q_1-q_5)+\frac{\sqrt3}{2}(q_2-q_6),\qquad
s_3=-\frac12(q_1-q_3)+\frac{\sqrt3}{2}(q_2-q_4).
$$
Consequently $V_2=(k/2)(s_1^2+s_2^2+s_3^2)$, with remainder $O(|q|^3)$. If $s=Bq$, then $\mathsf V=kB^TB$. Expanding the squares gives
$$
\boxed{\mathsf V=\frac k4\begin{pmatrix}
2&0&-1&\sqrt3&-1&-\sqrt3\\
0&6&\sqrt3&-3&-\sqrt3&-3\\
-1&\sqrt3&5&-\sqrt3&-4&0\\
\sqrt3&-3&-\sqrt3&3&0&0\\
-1&-\sqrt3&-4&0&5&\sqrt3\\
-\sqrt3&-3&0&0&\sqrt3&3
\end{pmatrix}.}
$$
The <small-oscillation mass and stiffness matrices> are $\mathsf T=mI_6$ and the displayed $\mathsf V$. Hence
$$
\boxed{L_2=\frac m2\sum_{j=1}^6\dot q_j^2-\frac12q^T\mathsf Vq,\qquad \mathsf Vv=m\omega^2v.}
$$
The two common translations give zero-mode vectors
$$
v_x=(1,0,1,0,1,0)^T,\qquad v_y=(0,1,0,1,0,1)^T.
$$
An infinitesimal counterclockwise rotation sends each equilibrium vertex $r_i$ to $Jr_i$, where $J(x,y)=(-y,x)$. Therefore a third zero-mode vector is
$$
\boxed{v_{\rm rot}=(-1,0,\tfrac12,\tfrac{\sqrt3}{2},\tfrac12,-\tfrac{\sqrt3}{2})^T,\qquad\omega_x=\omega_y=\omega_{\rm rot}=0.}
$$
Each vector makes all three $s_j$ vanish, so lies in the <kernel of a linear map> $B$ and therefore in the kernel of $\mathsf V$. The rows of $B$ are independent: the $q_4$ and $q_6$ entries first force the coefficients of its last two rows to vanish, then the first coefficient vanishes. Thus $\operatorname{rank}B=3$, and the <rank-nullity theorem> proves that these are exactly the three zero modes.
For radial motion take
$$
v_b=(0,1,\tfrac{\sqrt3}{2},-\tfrac12,-\tfrac{\sqrt3}{2},-\tfrac12)^T,
$$
so that $x_i(t)=(1+a(t))r_i$ to first order. The force on mass $i$ is $-ka\sum_{j\ne i}(r_i-r_j)=-3ka r_i$, since $\sum_i r_i=0$. Newton's equation therefore gives $m\ddot a=-3ka$. Equivalently $\mathsf Vv_b=3kv_b$. Hence the breathing <normal mode> has
$$
\boxed{\omega_b=\sqrt{\frac{3k}{m}}.}
$$
All three masses move radially outward together, then inward together, preserving the triangular shape. The remaining two <normal modes of an equilateral spring triangle> are shear deformations with squared frequency $3k/(2m)$; together with the breathing mode and three rigid zero modes they account for all six degrees of freedom.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-2-2010-ib-triangle-modes.png]
{title=Two translations, an infinitesimal rotation and the radial breathing mode of the equilateral spring triangle}
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