jmc

Let $M_1$, $M_2$, and $M_3$ be the midpoints of $AP$, $BP$, and $CP$, respectively. Then as a midline in triangle $PBC$, $M_2{M}_3$ is parallel to $BC$, and half the length of $BC$.



Since $G_3$ is the centroid of triangle $PAB$, $G_3$ divides median $A{M}_2$ in the ratio $2:1$. Similarly, $G_2$ divides median $A{M}_3$ in the ratio $2:1$. Therefore, triangles $A{G}_3{G}_2$ and $A{M}_2{M}_3$ are similar. Also, $G_2{G}_3$ is parallel to $M_2{M}_3$, and $G_2{G}_3$ is two-thirds the length of $M_2{M}_3$.

Therefore, $G_2{G}_3$ is parallel to $BC$, and $G_2{G}_3$ is one-third the length of $BC$. Likewise, $G_1{G}_2$ is parallel to $AB$, and $G_1{G}_2$ is one-third the length of $AB$. Hence, triangle $G_1{G}_2{G}_3$ is similar to triangle $ABC$, with ratio of similarity 1/3. The area of triangle $ABC$ is 18, so the area of triangle $G_1{G}_2{G}_3$ is $18\cdot (1/3{)}^2=\boxed{2}$.