= Solution
The <highest root> is $\theta=2\alpha_1+3\alpha_2$. In the <Extended Dynkin diagram>, the extra node $-\theta$ is long and is joined by a single bond to the long node $\alpha_1$; it is orthogonal to the short node $\alpha_2$. Deleting the short node therefore leaves the <A2 root system> <Dynkin diagram>. Equivalently, the six long <roots> are a closed <root subsystem>, so their <root spaces> together with the <Cartan subalgebra> form an <A2 root system> <Lie subalgebra>. The corresponding compact <subgroup> is $SU(3)\subset G_2$.
For the particular projection in the paper, choose the <A2 root system> <simple roots> as $\beta_1=\alpha_1+3\alpha_2$ and $\beta_2=\alpha_1$. Since these are long, $\beta_1^\vee=\alpha_1^\vee+\alpha_2^\vee$ and $\beta_2^\vee=\alpha_1^\vee$. Consequently the <Dynkin labels> restrict as
$$
\boxed{(a,b)^T\longmapsto P(a,b)^T=(a+b,a)^T,\qquad
P=\begin{pmatrix}1&1\\1&0\end{pmatrix}.}
$$
Projecting the six short <roots> gives $\pm(1,0),\pm(0,1),\pm(1,-1)$. Group these into the three <weights> of $\mathbf3$ and the three of $\overline{\mathbf3}$ listed in <3/a/solution>. The zero <weight> of $\mathbf7$ is left over, so
$$
\boxed{\mathbf7\downarrow_{SU(3)}=\mathbf3\oplus\overline{\mathbf3}\oplus\mathbf1.}
$$
The three positive long <roots> project as $(2,-3)\mapsto(-1,2)$, $(-1,3)\mapsto(2,-1)$ and $(1,0)\mapsto(1,1)$. Together with their negatives and the two zero <weights>, these are exactly the <Adjoint representation> $\mathbf8$. The remaining six short <roots> again give $\mathbf3\oplus\overline{\mathbf3}$. Hence the <branching of a representation> is
$$
\boxed{\mathbf{14}\downarrow_{SU(3)}=\mathbf8\oplus\mathbf3\oplus\overline{\mathbf3}.}
$$
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