 Solution

Consider one normalized measurement vector $|u⟩=a|0⟩+b|1⟩$ on C. The unnormalized partial inner product on AB isIts tensor-product coefficient matrix and determinant areThe two-qubit product-state determinant criterion therefore says that this branch is entangled whenever $a\ne 0$. 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 $\{|u_0⟩,|u_1⟩\}$, completeness gives $|⟨0|u_0⟩{|}^2+|⟨0|u_1⟩{|}^2=1$. At least one vector has $a\ne 0$. Hence 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.