 Solution

Order coordinates as $q=(x_1,x_2,x_3;y_1,y_2,y_3)$. For either uniform displacement, $t_x=(1,1,1;0,0,0)$ or $t_y=(0,0,0;1,1,1)$, every relative displacement and every extension vanishes. These normal modes have zero normal frequency and represent two independent planar translations.

For $r=(-1,-1,2;\sqrt{3},-\sqrt{3},0)$, substitution givesThus $Kr=0$ too. A small rigid rotation through angle $\alpha$ moves equilibrium position $(X_i,Y_i)$ by $\alpha (-Y_i,X_i)$. With the specified positions this displacement is $-\alpha r/2$, identifying the third vector as infinitesimal rigid rotation. HenceA zero-frequency normal coordinate obeys $\ddot{Q}=0$, so $Q=A+Bt$. This permits a constant displacement or a uniform velocity at the level of the linearized equations. In particular, the rotational pattern preserves lengths to first order; finite-angle rotation follows a curved path outside a single linear displacement pattern.