 Solution

At the south pole $\mathbf{r}=-R\hat{z}$. Choose plate C as the instantaneous reference frame. Since $R=4\times 10^9$ mm, the two components of relative angular speed give 20 and 48 mm/year. The plate velocities in the fixed tangent-plane axes areThe plate velocity triangle therefore has sidesThe physical boundary rays point along $+x$ for AB, $+y$ for AC and $(12,-5)$ for BC. Plate A occupies the quadrant between the first two rays, B the narrow wedge between AB and BC, and C the remainder. The stated overriding relations make AB and AC subduction trenches, while BC remains a transform fault: this is a trench-trench-transform triple junction. Put the subduction teeth on B's side of AB and A's side of AC, identifying B above A and A above C.

For kinematic stability there must be a single junction velocity $\mathbf{U}$ compatible with all three boundary-normal motions. Use the standard idealization that each trench's normal velocity is that of its overriding plate, with no independent rollback. AB is parallel to $x$, so $U_y=v_{B,y}=20$. AC is parallel to $y$, so $U_x=v_{A,x}=-48$. The transform requires $\mathbf{U}-{\mathbf{v}}_C$ to be parallel to $(12,-5)$, which is true for this intersection. HenceThe southern triple junction is instantaneously kinematically stable. It is stationary relative to B, migrates along AB relative to A, and migrates along BC relative to C. This compatibility result does not promise long-term stability if plate rotations or trench rollback change. Notice that the three relative velocities close even though the overriding order is not a statement that all three plates move toward one common stationary point.