= Solution
Write $h_{ij}$ for the spatial <metric tensor> and lower the <shift vector> with $h_{ij}$. For the negative-shift convention, the four-dimensional metric and its inverse have components
$$
g_{00}=-N^2+N_iN^i,\quad g_{0i}=-N_i,\quad g_{ij}=h_{ij},\qquad g^{00}=-N^{-2},\quad g^{0i}=-N^iN^{-2}.
$$
The normal has covariant components $n_\mu=(-N,0,0,0)$, so $n_\mu n^\mu=-1$ and it annihilates every tangent vector to the spatial slice. Since $n_i=0$, its <covariant derivative> gives $n_{i;j}=N\Gamma^0{}_{ij}$ and therefore $K_{ij}=-N\Gamma^0{}_{ij}$.
Compute the required <Christoffel symbol> using the symmetric connection formula. Its time-index term and spatial-index term combine to give
$$
\Gamma^0{}_{ij}=\frac{1}{2N^2}\left(\dot h_{ij}+\partial_iN_j+\partial_jN_i-N^k(\partial_i h_{jk}+\partial_jh_{ik}-\partial_kh_{ij})\right).
$$
The spatial expression in parentheses is $\dot h_{ij}+D_iN_j+D_jN_i$, where $D$ is the <covariant derivative> of $h$. Thus the <extrinsic curvature with a negative shift> is
$$
\boxed{K_{ij}=-\frac{1}{2N}(\dot h_{ij}+D_iN_j+D_jN_i).}
$$
This derivation fixes both signs from the actual <3+1 decomposition of spacetime>, rather than transferring a formula using the opposite <shift vector> convention. The connection formula used here has $g_{\lambda\kappa,\nu}$ as its second differentiated term; the printed repeated-index version is a typographical error.
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