= Solution
Suppose the inclined population samples its arguments of periapsis and nodal orientations, or that sufficiently slow precession supplies this sampling. Near the <planet> the vertical amplitude is $H\simeq a_pI$. For uniformly distributed vertical phase, $z=H\sin\psi$ has <probability density function> $[\pi\sqrt{H^2-z^2}]^{-1}$, so the midplane volume probability is $\Sigma_c/(\pi H)$.
In the genuinely three-dimensional regime $R_p\ll H\ll a_p$, the <geometric collision cross-section> is $\pi R_p^2$. The <relative velocity> differs from the coplanar value only at order $I^2$ for the specified high-speed encounters. Therefore the <inclined comet collision time> is
$$
\boxed{t_{\rm col,I}\simeq\frac{H}{R_p^2\Sigma_c(a_p)v_{\rm rel}}\simeq\frac{\pi^2I Q^{3/2}a_p^2\sqrt{1-q/a_p}}{R_p^2\sqrt{GM_\star}\sqrt{3-2\sqrt{2q/a_p}}}.}
$$
At fixed $\rho_p,M_\star,I$ the leading scaling is $Q^{3/2}M_p^{-2/3}a_p^2$, with no leading $q$ dependence when $q\ll a_p$. The displayed factor retains the first dependence on $q/a_p$. Also $t_{\rm col,I}/t_{\rm col}\simeq2a_pI/R_p$ in this averaging convention.
\b[Inclination alone does not specify a collision time for one fixed orbit.] If its <orbital nodes> miss the <planet>'s orbit, a tilted <Kepler orbit> can have no collisions at all. The finite result is an orientation-averaged population result, or requires precession or scattering to sample those orientations. When $a_pI\lesssim R_p$, the three-dimensional estimate crosses over to the coplanar result instead of tending to zero.
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