Undetectable cheater in Shamir's Secret Sharing scheme
The way to prove this is to follow the same proof that Shamir's secret sharing is perfectly secret. Specifically, given any two points all secrets are possible since there is a polynomial going through every possible secret and the two given points. Since the polynomial is random, all of these polynomials have the same probability. The same thing is happening here: given any set of three points, there is a polynomial going through those three points. Now, assume that you know exactly what the secret is. If you look at any subset of two of the three points, there is a polynomial going through those two points and the secret that you know. Thus, it is impossible to know which of the three parties is cheating, even if you actually know the secret you are supposed to get.