$x + \frac1y = y + \frac1z = z + \frac1x$, then value of $xyz$ is? If $x,y,z$ are distinct positive numbers, such that $$x + \frac1y = y + \frac1z = z + \frac1x $$ then value of $xyz$ is?
$$A)\ 4\quad  B)\ 3\quad C)\ 2\quad  D)\ 1$$
My attempt: 
1.I equaled the equation to '$k$'. Using the AM-GM inequality, I found that $k>2$ (the equality does not hold because all are distinct). However, this, I couldn't put to much use.
2. For the next attempt, I substituted the values of $x$ and $y$ in terms of $z$ and $k$ in $xyz$. What I am getting is $xyz=zk^2 - k - z$. That's the farthest I could do..Please, help.

Other answers have shown that the question is flawed. For fun, let's remove the condition that $x,y,z$ are positive, but keep the requirement that $x,y,z$ are distinct, and try to find all solutions $(x,y,z)$.
Trivially, $x,y,z \neq 0$. If we set $x+\dfrac{1}{y} = y+\dfrac{1}{z} = z + \dfrac{1}{x} = k$, for som real number $k$, then 
\begin{align}
x &= k - \dfrac{1}{y}
\\
y &= k - \dfrac{1}{z}
\\
z &= k - \dfrac{1}{x}
\end{align}
If we substitute equation (3) into (2) and then substitute that into (1), we get $$x = k-\dfrac{1}{k-\dfrac{1}{k-\tfrac{1}{x}}}.$$
By unraveling this fraction, we eventually get the equation $$(k^2-1)(x^2-kx+1) = 0.$$
If $k \neq \pm 1$, then the solutions $x$ will satisfy $$x^2-kx+1 = 0 \iff x = k - \dfrac{1}{x} = z,$$ and we won't have distinctness. 
If $k = 1$, then $z = 1-\dfrac{1}{x} = \dfrac{x-1}{x}$ and $y = 1-\dfrac{1}{z} = -\dfrac{1}{x-1}$. 
So the solutions in this case are $(x,y,z) = \left(x,-\dfrac{1}{x-1},\dfrac{x-1}{x}\right)$, all of which satisfy $xyz = -1$.
If $k = -1$, then $z = -1-\dfrac{1}{x} = -\dfrac{x+1}{x}$ and $y = -1-\dfrac{1}{z} = -\dfrac{1}{x+1}$.
So the solutions in this case are $(x,y,z) = \left(x,-\dfrac{1}{x+1},-\dfrac{x+1}{x}\right)$, all of which satisfy $xyz = 1$.
Therefore, the only solutions where $x,y,z$ are distinct are all of the form $(x,y,z) = \left(x,-\dfrac{1}{x-1},\dfrac{x-1}{x}\right)$ for $x \neq 0,1$ or $(x,y,z) = \left(x,-\dfrac{1}{x+1},-\dfrac{x+1}{x}\right)$ for $x \neq 0,-1$. (It's easy to check that these satisfy the distinctness condition.) Also, the only possible values of $xyz$ are $\pm 1$.
