 Solution

The corresponding conditional vectors are $|0⟩/\sqrt{2}$ and $|+⟩/\sqrt{2}$. The second has a nonzero $|1⟩$ component while the first does not, so they are linearly independent. Therefore the third state is also entangled. Orthogonality of these two conditional vectors is not required for this conclusion.

For an independent matrix check, expanding the second factor givesThe original expression is normalized because its two first-wire factors are orthogonal: its squared norm is $(1+1)/2=1$, despite the overlap between $|0⟩$ and $|+⟩$ on the second wire.