1. Ghost rows c\,c\, c and c\,c\,K : diagnosis and fix

Q05 reported both rows OPEN: the assembled one-loop row did not vanish and no rescaling of the four channels closed it. Q06 splits every channel by orientation of the two identical ghosts and fits all six coefficients freely on the 4D (non-evanescent) part. The fit has a unique solution (up to scale), every coefficient $\pm 1$: the h channel and both c channels enter as assembled in Q05, but the K channel $\delta S_1/\delta h{|}_{X{c}_j}\cdot {\Gamma }_{K{c}_i}$ enters with the opposite orientation sign. Reason: in the K channel the external $c_i$ sits inside ${\Gamma }_{K{c}_i}$, to the right of $c_j$, while in the h channel the order is reversed; bringing the monomial to the row ordering $c_1{c}_2$ costs one Grassmann transposition. The tree row has no K channel, so the tree calibration (${\sigma }_c=-1$ for $\bar{c}$, $+1$ for K) is unchanged. With this fix the 4D part of both rows vanishes exactly, and every single-channel mutation (drop $h_{1,2}$, $K_{1,2}$, $c_1$, $c_2$, flip ${\sigma }_c$, restore the Q05 K sign) leaves a nonzero 4D residual.

Second route for G10. The $c^{\ast }cc$ triangle was recomputed with a re-routed loop momentum (graviton carries $l-p_1$) and the two $c$ slots of $V_{c^{\ast }cc}$ entered in the opposite order (Grassmann sign $-1$): bit-identical to the Q05 pole. G10 was therefore not the defect.

What remains. A purely evanescent remainder (22 terms for $ccK$, 24 for $cc\bar{c}$; every term carries one $T$ between external polarisations). Because the row is linear and its 4D part vanishes, local evanescent counterterms equal to minus the evanescent pole parts (p22 prescription) remove it; whether those counterterms are $s_0$-consistent on the full ${\kappa }^3$ evanescent domain is OPEN; it is recorded explicitly (sha256 in the certificate) and is not called an obstruction.