= Solution
For the stated geometry and numerical masses, the inertia and <stiffness matrices> are
$$
M=\operatorname{diag}\left(2,\frac14,\frac2{25}\right),\qquad
K=\begin{pmatrix}5&-2&0\\-2&3&-1\\0&-1&1\end{pmatrix}.
$$
A <normal mode> $\boldsymbol\theta=\mathbf v e^{i\omega t}$ requires $(K-\omega^2M)\mathbf v=0$. Writing $z=\omega^2$, its <characteristic equation> is
$$
\det(K-zM)=-\frac{4z^3-108z^2+613z-600}{100}=0.
$$
At the claimed $z=6$, this determinant is $-27/50\ne0$. Hence \b[the printed claim that $\sqrt6$ is a normal frequency is false for the stated data], and there is no corresponding nonzero <normal mode>.
The complete geometry-based answer is obtained from the three positive roots $z_j$:
$$
\boxed{\begin{aligned}
4z_j^3-108z_j^2+613z_j-600&=0,\\
(\omega_{\rm low},\omega_{\rm middle},\omega_{\rm high})
&\simeq(1.111471775,\ 2.491790287,\ 4.422172730).
\end{aligned}}
$$
An exact closed form for the squared <angular frequencies> is
$$
z_j=9+\sqrt{\frac{359}{3}}\cos\left[\frac13\arccos\frac{2745\sqrt3}{359^{3/2}}-\frac{2\pi j}{3}\right],\qquad j=0,1,2.
$$
They are approximately $19.555611657$, $6.209018836$ and $1.235369507$, respectively. For any of these roots, a corresponding <eigenvector> can be chosen as
$$
\boxed{\mathbf v(z)=\begin{pmatrix}
\tfrac12\left[(3-z/4)(1-2z/25)-1\right]\\
1-2z/25\\
1
\end{pmatrix}.}
$$
The last two rows of $(K-zM)\mathbf v=0$ give this expression directly; the first is the characteristic equation. These three modes span all motion, with arbitrary sine and cosine amplitudes. Positivity of $M$ and of the spring <potential energy> ensures that all squared <angular frequencies> are positive.
If instead one takes the PDF's printed $I_1=m_1a^2/6$ as a stipulated inertia, then $M=\operatorname{diag}(1/2,1/4,2/25)$ and
$$
\det(K-zM)=-\frac{2z^3-69z^2+626z-1200}{200}.
$$
The corresponding <angular frequencies> are approximately
$$
\boxed{(1.616170898,\ 3.315716400,\ 4.570997307).}
$$
The same formula for $\mathbf v(z)$ applies, because its derivation used only the second and third rows. Here the determinant at $z=6$ is $-63/25\ne0$, so following the printed inertia does not repair the frequency claim. The PDF therefore needs a further correction to its numerical data or asserted frequency; the intended correction cannot be inferred uniquely.
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