Compute the output of the quantum teleportation circuit

Since your circuit is teleportation, $|Crangle =|varphirangle$ and since you measured $|11rangle$ on $|ABrangle$ the answer is $|ABCrangle = |11rangle|varphirangle$.

Now, let look why this is true.

Firstly Hadamard and CNOT gate on second and third qubit prepares entangled Bell state $|beta_{00}rangle = frac{1}{sqrt{2}}(|00rangle + |11rangle)$.

Now the circuit is in state

$$ |varphirangle|beta_{00}rangle = frac{1}{sqrt{2}}(alpha|0rangle+beta|1rangle)(|00rangle + |11rangle) = frac{1}{sqrt{2}}[alpha|0rangle(|00rangle + |11rangle) + beta|1rangle(|00rangle + |11rangle)] $$

Then you apply CNOT controlled by first qubit and targeting second qubit. This will negate second qubit in case the first qubit is in state $|1rangle$. This means that only part $beta|1rangle(|00rangle + |11rangle)$ is influenced.

Now, state of the circuit is changed to

$$ frac{1}{sqrt{2}}[alpha|0rangle(|00rangle + |11rangle) + beta|1rangle(|10rangle + |01rangle)] $$

Application of Hadamard gate on the first qubit change the state further to

$$ frac{1}{2}[alpha(|0rangle + |1rangle)(|00rangle + |11rangle) + beta(|0rangle - |1rangle)(|10rangle + |01rangle)] $$

because $H|0rangle = frac{1}{sqrt{2}}(|0rangle + |1rangle)$ and $H|1rangle = frac{1}{sqrt{2}}(|0rangle - |1rangle)$.

Since you will measure first and second qubit, it is convenient to rearange the state to separate first two qubits. So, you can rewrite the state as

$$ frac{1}{2} big( |00rangle(alpha|0rangle + beta|1rangle) + |01rangle (alpha|1rangle + beta|0rangle) + |10rangle (alpha|0rangle - beta|1rangle) + |11rangle (alpha|1rangle - beta|0rangle) big) $$

In your case you measured $|11rangle$ on first and second qubit. This means that the third qubit is in state

$$ (alpha|1rangle - beta|0rangle) $$

Since both first and second qubits are in state $|1rangle$ both CNOT gates after measurement will be activated. The first one change the state of third qubit to

$$ (alpha|0rangle - beta|1rangle) $$

Next two Hadamards together with CNOT implements controlled $Z$ gate which change a phase to oposite in case input qubit is in state $|1rangle$. This leads to final state of third qubit

$$ (alpha|0rangle + beta|1rangle) $$

Hence, you can see that state from first qubit was teleported to third qubit.

Note: based on Nielsen and Chuang, pg. 27 and expanded