= Solution
Consider one normalized measurement <vector> $|u\rangle=a|0\rangle+b|1\rangle$ on C. The unnormalized <partial inner product> on AB is
$$
|\chi_u\rangle=(I_{AB}\otimes\langle u|)|W\rangle=\frac1{\sqrt3}\bigl(\overline b|00\rangle+\overline a|01\rangle+\overline a|10\rangle\bigr).
$$
Its <tensor-product coefficient matrix> and determinant are
$$
C_u=\frac1{\sqrt3}\begin{pmatrix}\overline b&\overline a\\\overline a&0\end{pmatrix},\qquad \det C_u=-\frac{\overline a^{\,2}}3.
$$
The <two-qubit product-state determinant criterion> therefore says that this branch is entangled whenever $a\ne0$. Its <probability> is $(|b|^2+2|a|^2)/3>0$, so such a branch is physically possible and can be normalized without changing whether its determinant vanishes.
For an <orthonormal basis> $\{|u_0\rangle,|u_1\rangle\}$, completeness gives $|\langle0|u_0\rangle|^2+|\langle0|u_1\rangle|^2=1$. At least one <vector> has $a\ne0$. Hence \b[at least one nonzero-probability outcome leaves AB entangled], proving <conditional entanglement of the three-qubit W state>. Some bases give two entangled branches; the <computational basis> gives one entangled and one product branch. The claim is not that every outcome must entangle AB.
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