= Solution
At <latitude> $\phi$ and <longitude> $\lambda$, $\mathbf r=R(\cos\phi\cos\lambda,\cos\phi\sin\lambda,\sin\phi)$. Motion is $\mathbf v_{B/C}=\boldsymbol\omega_{B/C}\times\mathbf r$. A meridian boundary has horizontal normal $\hat e_\lambda=(-\sin\lambda,\cos\lambda,0)$, so its across-boundary <velocity> is
$$
\mathbf v_{B/C}\cdot\hat e_\lambda=-R\sin\phi\,(5\cos\lambda+12\sin\lambda)\times10^{-9}\,\mathrm{yr^{-1}}.
$$
For a <transform fault> this must vanish along the whole meridian, not only at an isolated point. Thus $5\cos\lambda+12\sin\lambda=0$. Select the solution in the western hemisphere:
$$
\boxed{\lambda=-22.620^\circ=22.620^\circ\mathrm W.}
$$
The other solution is the antipodal meridian in the eastern hemisphere. Geometrically, the transform meridian is in a great-circle plane perpendicular to the <Euler angular velocity vector>. Its outgoing tangent direction at either pole, expressed in the fixed $x,y$ axes, is $\hat t_F=(12,-5)/13$.
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