= Solution
The gravitational three-body <Lagrangian> is
$$
L
=\frac12\sum_{i=1}^3m_i|\dot{\mathbf r}_i|^2
+G\left(
\frac{m_1m_2}{|\mathbf r_1-\mathbf r_2|}
+\frac{m_1m_3}{|\mathbf r_1-\mathbf r_3|}
+\frac{m_2m_3}{|\mathbf r_2-\mathbf r_3|}
\right).
$$
Put $M_{12}=m_1+m_2$ and $M=M_{12}+m_3$. Solving the definitions of the <Jacobi coordinates for three particles> gives
$$
\begin{aligned}
\mathbf r_1&=\mathbf c+\frac{m_3}{M}\mathbf b
+\frac{m_2}{M_{12}}\mathbf a,\\
\mathbf r_2&=\mathbf c+\frac{m_3}{M}\mathbf b
-\frac{m_1}{M_{12}}\mathbf a,\\
\mathbf r_3&=\mathbf c-\frac{M_{12}}M\mathbf b.
\end{aligned}
$$
Substitution into the kinetic energy makes all cross terms cancel:
$$
\boxed{
T
=\frac12\alpha|\dot{\mathbf a}|^2
+\frac12\beta|\dot{\mathbf b}|^2
+\frac12\gamma|\dot{\mathbf c}|^2
},
$$
where
$$
\boxed{
\alpha=\frac{m_1m_2}{m_1+m_2},
\qquad
\beta=\frac{(m_1+m_2)m_3}{m_1+m_2+m_3},
\qquad
\gamma=m_1+m_2+m_3
}.
$$
The pair separations are
$$
\mathbf r_1-\mathbf r_2=\mathbf a,
\qquad
\mathbf r_1-\mathbf r_3
=\mathbf b+\frac{m_2}{M_{12}}\mathbf a,
\qquad
\mathbf r_2-\mathbf r_3
=\mathbf b-\frac{m_1}{M_{12}}\mathbf a.
$$
Hence the <potential energy>[gravitational potential energy] is
$$
V(\mathbf a,\mathbf b)
=-\frac{Gm_1m_2}{|\mathbf a|}
-\frac{Gm_1m_3}
{\left|\mathbf b+\frac{m_2}{M_{12}}\mathbf a\right|}
-\frac{Gm_2m_3}
{\left|\mathbf b-\frac{m_1}{M_{12}}\mathbf a\right|},
$$
which is independent of $\mathbf c$. Thus $\mathbf c$ is an <ignorable coordinate> and
$$
\frac d{dt}(\gamma\dot{\mathbf c})=0,
\qquad
\boxed{\ddot{\mathbf c}=0}.
$$
The center of mass moves uniformly because the isolated system has no external force.
Finally, substituting the inverse coordinate transformation into
$$
\mathbf L=\sum_i m_i\mathbf r_i\times\dot{\mathbf r}_i
$$
again cancels every cross term and yields
$$
\boxed{
\mathbf L
=\alpha\mathbf a\times\dot{\mathbf a}
+\beta\mathbf b\times\dot{\mathbf b}
+\gamma\mathbf c\times\dot{\mathbf c}
}.
$$
This is the <angular momentum decomposition in three-body Jacobi coordinates>.
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