The domain over which we solve the equation is $\mathbb{R} \setminus \{2,6\}$.
We can now cross-multiply to get rid of the fractions, we get $(x-1)(x-6)=(x-k)(x-2)$.
Simplifying that, we get $7x-6 = (k+2)x - 2k$. Clearly for $k=5$ we get the equation $-6=-10$ which is never true. The answer is $\boxed{\mathrm{ (E)}\ 5}$
For other $k$, one can solve for $x$: $x(5-k) = 6-2k$, hence $x=\frac {6-2k}{5-k}$. We can easily verify that for none of the other 4 possible values of $k$ is this equal to $2$ or $6$, hence there is a solution for $x$ in each of the other cases.
-Edited by XxHalo711 (typo within the solution)