One-time pad, zero key and Shannon
Well, for perfect secrecy, we require that for all message distributions over $\mathcal{M}$, all messages $m\in\mathcal{M}$, and all (possible) ciphertexts $c$ it holds that

$$Pr[M=m\ |\ C = c] = Pr[M=m]$$

In particular, that means, if we can find a single counterexample, i.e. a distribution over $\mathcal{M}$, a message $m\in\mathcal{M}$, and a ciphertexts $c$ such that

$$Pr[M=m\ |\ C = c] \neq Pr[M=m]$$

then, by our definition, the encryption scheme does not offer perfect secrecy.

So, let's take the uniform distribution over $\mathcal{M}$. For this distribution, we obviously have $$Pr[M=c]=1/|\mathcal{M}|.$$
Now, given that $0^n$ is not a valid key, what is the probability  $$Pr[M=c\ |\ C = c]?$$ Are the two probabilities equal?