= Solution
The <no-cloning theorem> states that no fixed deterministic <quantum operation> can perfectly duplicate every unknown <pure state>. For a <unitary> copier with a fixed blank, suppose it cloned two normalized vectors $|a\rangle,|b\rangle$. Preservation of <inner products> would give $\langle a|b\rangle=\langle a|b\rangle^2$, impossible for a nonorthogonal distinct pair with overlap strictly between zero and one in modulus. Including an environment does not help: its final overlap has modulus at most one, so one would require $|\langle a|b\rangle|\leq|\langle a|b\rangle|^2$. Thus a universal perfect copier is impossible. <Orthogonal> known alternatives do not contradict this restriction.
Label the three <qubits> as input $S$, Alice's entangled <qubit> $A$ and Bob's <qubit> $B$. Initially their joint state is
$$
\frac1{\sqrt2}[\alpha_0(|000\rangle+|011\rangle)+\alpha_1(|100\rangle+|111\rangle)].
$$
The <controlled-NOT gate> from $S$ to $A$ changes this to
$$
\frac1{\sqrt2}[\alpha_0(|000\rangle+|011\rangle)+\alpha_1(|110\rangle+|101\rangle)].
$$
Apply $H|0\rangle=(|0\rangle+|1\rangle)/\sqrt2$ and $H|1\rangle=(|0\rangle-|1\rangle)/\sqrt2$ to the first <qubit> and group by Alice's two computational-basis outcomes. The result is
$$
\boxed{\frac12\bigl[|00\rangle|\psi\rangle+|01\rangle X|\psi\rangle+
|10\rangle Z|\psi\rangle+|11\rangle XZ|\psi\rangle\bigr].}
$$
In expanded form the last two Bob factors are $\alpha_0|0\rangle-\alpha_1|1\rangle$ and $\alpha_0|1\rangle-\alpha_1|0\rangle$. \b[Their Alice labels must be $10$ and $11$; the repeated $00$ labels in the source are printing errors.]
Alice measures her two <qubits> and sends the two classical bits $(a,b)$ to Bob. Each outcome has probability $1/4$ and Bob's conditional state is $X^bZ^a|\psi\rangle$. He applies $\boxed{Z^aX^b}$ to recover the input, up to physically irrelevant <global phase>. Explicitly the corrections for $00,01,10,11$ are $I,X,Z,ZX$. This is <quantum teleportation>: Alice's input is consumed by <measurement>, so no second surviving copy is created. Before her classical message, Bob's <reduced density operator> is $I/2$, independent of the input, so the protocol also supplies no faster-than-light communication.
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