= Solution
Let $\hat x$ point to the equatorial meridian at zero <longitude>, $\hat y$ to the equator at $90^\circ$ E, and $\hat z$ to the north pole. Use a right-hand positive <Euler angular velocity vector>. Relative <rotations> add as <vectors>, not as geographic coordinates:
$$
\boldsymbol\omega_{B/C}=\boldsymbol\omega_{B/A}+\boldsymbol\omega_{A/C}=(5\hat x+12\hat y)\times10^{-9}\,\mathrm{rad\,yr^{-1}}.
$$
Consequently
$$
\boxed{|\boldsymbol\omega_{B/C}|=13\times10^{-9}\,\mathrm{rad\,yr^{-1}},\qquad (\phi,\lambda)=(0^\circ,67.380^\circ\mathrm E).}
$$
Here $\lambda=\tan^{-1}(12/5)$ is the <longitude> of the positive <Euler pole>, and its <latitude> $\phi$ is zero. The antipodal pole with the opposite angular-velocity sign is an equivalent convention; mixing a pole reversal with an unchanged sign would give the wrong motion.
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