= 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 $r=(-1,-1,2;\sqrt3,-\sqrt3,0)$, substitution gives
$$
e_{12}=0,\qquad e_{13}=\tfrac32-\tfrac32=0,\qquad e_{23}=-\tfrac32+\tfrac32=0.
$$
Thus $Kr=0$ 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>. Hence
$$
\boxed{\omega^2(t_x)=\omega^2(t_y)=\omega^2(r)=0.}
$$
A 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.
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