Figure 1.16: FANOUT with the Toffoli gate, with the second bit being the input to the FANOUT (and the other two bits standard ancilla states), and the output from the FANOUT appearing on the second and third bits.
Source: Quantum Computation and Quantum Information: 10th Anniversary Edition (Fig 1.16, p.30)
by Michael A. Nielsen & Isaac L. Chuang
How can we say that the third qubit is in the state $a$?
Let's assume $a = x|0\rangle + y|1\rangle$. We know that the output of the bottom two qubits is $x|00\rangle+y|11\rangle$ (entangled qubits) and the output of the second qubit must be $x|0\rangle+y|1\rangle$. Which says that they both have to be in the state $|0\rangle$ and $|1\rangle$ at the same time, and so the third qubit should have the same probabilities as like of the second qubit. But it can be $-x|0\rangle + y|1\rangle$ and many more possibilities as $x$ and $y$ are complex (as probabilities are square of modulo of coefficients).
Please correct me if my interpretation and understanding is wrong.