= Solution
For three unit <masses>, $M=I$ and
$$
K=k\begin{pmatrix}3&-1&-1\\-1&3&-1\\-1&-1&3\end{pmatrix}
=k(4I-\mathbf1\mathbf1^T).
$$
The uniform vector has $\mathbf1\mathbf1^T\mathbf1=3\mathbf1$, whereas every vector with coordinate sum zero is annihilated by $\mathbf1\mathbf1^T$. One orthonormal choice of <normal modes> is therefore
$$
\boxed{\begin{aligned}
\mathbf v_1&=\frac1{\sqrt3}(1,1,1)^T,&\omega_1&=\sqrt{k},\\
\mathbf v_2&=\frac1{\sqrt2}(1,-1,0)^T,&\omega_2&=2\sqrt{k},\\
\mathbf v_3&=\frac1{\sqrt6}(1,1,-2)^T,&\omega_3&=2\sqrt{k}.
\end{aligned}}
$$
If $k=1$ is retained from (b), the <angular frequencies> are $1,2,2$. The equal upper <frequencies> follow from <degenerate normal modes from permutation symmetry>: the stiffness is scalar on the two-dimensional sum-zero subspace. In the uniform mode the inter-bar strings do not change length, but all three beam connections stretch, giving the nonzero lower <frequency>. The motion is an arbitrary sine-cosine <superposition> of these modes about the gravitational <equilibrium>.
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