= Solution
The <kinetic energy> is $K=\tfrac12m(3\dot x^2+3\dot y^2+4\dot z^2)$. Differentiating the given potential and applying <Newton's second law> gives
$$
\boxed{3\ddot x+\Omega^2(2x-z)=0,\quad3\ddot y+\Omega^2(2y-z)=0,\quad4\ddot z+\Omega^2(2z-x-y)=0.}
$$
These are the <normal modes of a symmetric three-mass string>, with mass matrix $M=m\operatorname{diag}(3,3,4)$ and stiffness matrix $m\Omega^2\begin{pmatrix}2&0&-1\\0&2&-1\\-1&-1&2\end{pmatrix}$. The coordinate order here is $(x,y,z)$, although the spatial order along the string is $x,z,y$.
Reflection symmetry splits the <generalized eigenvalue problem> into an antisymmetric mode $x=-y$, $z=0$, with $\omega^2=2\Omega^2/3$, and symmetric modes $x=y=q$. Writing $\mu=\omega^2/\Omega^2$, these obey $z=(2-3\mu)q$ and $(2-4\mu)z=2q$. Thus $(2-4\mu)(2-3\mu)-2=2(6\mu-1)(\mu-1)=0$. The three results are
$$
\boxed{\begin{array}{c|c}\omega&x:y:z\\\hline\Omega/\sqrt6&2:2:3\\\Omega\sqrt{2/3}&1:-1:0\\\Omega&1:1:-1\end{array}.}
$$
The displacement vectors are mutually orthogonal under the mass-weighted inner product $u^TMv$, as required for these conservative <normal modes>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-2-three-mass-modes.png]
{title=The three transverse normal modes in their physical order along the string}
{description=Original mode shapes with segment length $a=1$. The moving particles have masses $3m,4m,3m$ from left to right and positions $x,z,y$. Vertical displacements are normalized independently in each panel; straight string segments join the particles and fixed endpoints.}
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