= Solution
The pinned third bar has zero displacement, so remove its row and column from the <stiffness matrix> while retaining the <springs> that connect the moving bars to it. The remaining unit-mass system has
$$
K_{\rm pin}=k\begin{pmatrix}3&-1\\-1&3\end{pmatrix}.
$$
Its <eigenvectors> $(1,1)^T$ and $(1,-1)^T$ have <eigenvalues> $2k$ and $4k$. Hence
$$
\boxed{\omega_{\rm in\ phase}=\sqrt{2k},\qquad\omega_{\rm out\ of\ phase}=2\sqrt{k}.}
$$
For $k=1$ these are $\sqrt2$ and $2$. The fixed value $X_3=0$ changes the static offsets and reaction force but not this linearized <matrix>.
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