jmc

We label the points as following: the centers of the circles of radii $3,6,9$ are $O_3,O_6,O_9$ respectively, and the endpoints of the chord are $P,Q$. Let $A_3,A_6,A_9$ be the feet of the perpendiculars from $O_3,O_6,O_9$ to $\overline{PQ}$ (so $A_3,A_6$ are the points of tangency). Then we note that $\overline{O_3{A}_3}\parallel \overline{O_6{A}_6}\parallel \overline{O_9{A}_9}$, and $O_6{O}_9:O_9{O}_3=3:6=1:2$. Thus, $O_9{A}_9=\frac{2\cdot O_6{A}_6+1\cdot O_3{A}_3}{3}=5$ (consider similar triangles). Applying the Pythagorean Theorem to $△O_9{A}_{9}P$, we find that$$P{Q}^2=4(A_{9}P{)}^2=4[(O_{9}P{)}^2-(O_9{A}_9{)}^2]=4[9^2-5^2]=\boxed{224}$$