Desrcibing all the equivalent martingale measures (EMM) Exercise :

Consider the finite sample space $\Omega = \{\omega_1, \omega_2, \omega_3\}$ and the probability space $(\Omega, \mathcal{F}, \mathbb P)$ with $\mathcal F:= 2^\Omega$ and let $\mathbb P$ be a probability measure such that $\mathbb P[\{\omega_1\}] > 0$ for all $i=1,2,3$ . 
Let :
  $$\bar{S}_0 = \begin{pmatrix} 1\\2\\7 \end{pmatrix}, \quad\bar{S}_1(\omega_1) = \begin{pmatrix} 1\\3\\9\end{pmatrix}, \quad \bar{S}_1(\omega_2) = \begin{pmatrix} 1\\1\\5\end{pmatrix}, \quad \bar{S}_1(\omega_3) = \begin{pmatrix} 1\\13\\10 \end{pmatrix}$$
  Describe all the equivalent martingale measures.

Definitions and request for help/elaboration :

Theorem : Let $\mathbb P, \mathbb Q$ be two probability measures  over the space $(\Omega, \mathcal{F})$. Then, it is $\mathbb P \sim \mathbb Q$ if and only if there exists a random variable $X>0, \; X \in L^1(\mathbb P)$ such that :
  $$\mathbb E_\mathbb Q[\mathbf{1}_A] = \mathbb Q(A) = \int_AX\mathrm{d}\mathbb P=\mathbb E_\mathbb P[X\mathbf{1}_A], \quad \forall A \in \mathcal F$$
Definition : A probability measure $\mathbb Q$ over the space $(\Omega, \mathcal{F})$ is called a martingale measure (MM) if it is :
  $$S_1 \in L^1(\mathbb Q) \quad \text{and} \quad S_0 = \mathbb{E}_\mathbb Q\bigg[\frac{S_1}{1+r}\bigg]$$
Definition : If $\mathbb Q$ is a martingale measure and it also is $\mathbb Q \sim \mathbb P$, then $\mathbb Q$ is called an equivalent martingale measure (EMM).
Definition : We define $\mathcal{P}$ to be the set of all the equivalent martingale measures :
$$\mathcal{P} = \{\mathbb Q|\mathbb Q \sim \mathbb P, \mathbb Q  \; \text{is an equivalent martingale measure}\}$$

Question : Now, given these definitions, what is needed for me to carry out so I can answer properly to the exercise given ? What am I expected to do and how ? I would really appreciate a thorough elaboration as I haven't been properly introduced to martingales and measure theory yet (there are just tools needed for a higher-level class).
 A: From the fact that the first asset does not change value, we can infer that the interest rate is zero. The problem is to find all probability measures $Q$ such that $$  S_0 = S_1Q(\omega_1) + S_2Q(\omega_2) +S_3Q(\omega_3),$$ and all of the three probabilities are greater than zero (and add up to one, of course). The equation expresses the fact that it is a martingale measure (the RHS is the expected discounted value). It needs all probabilities greater than zero in order to be equivalent to $P.$
As far as I can tell, there aren’t any (I find one probability measure that has the right expected value but it doesn’t have all probabilities nonzero.)
Edit
Going to expand on equivalence a bit. Two measures are equivalent if they have the same null events. I notice the part on equivalence is marked as a theorem, not a definition, so I suspect perhaps you were given this definition at at some point. Since the physical measure $P$ you've been given has $P(\omega)>0$ for all $\omega\in\Omega,$ the only null event is the empty set. Thus for $Q$ to be equivalent, it must also have the only null event be the emptyset, i.e. $Q(\omega)>0$ for all $\omega\in \Omega.$
To see this is equivalent to the theorem that there exists a random variable $X>0$ such that for any event $A$ $$ E_Q(1_A) = E_P(X1_A),$$ as I mentioned in the comments, just take $X(\omega) = \frac{Q(\omega)}{P(\omega)},$ for all $\omega$ such that $P(\omega) >0,$ and let it be defined as any positive number elsewhere. We see that the fact that P and Q are equivalent implies $X>0$. And we have (specializing to the sample space in the problem for clarity) $$ E_Q(1_A) = 1_A(\omega_1)Q(\omega_1) + 1_A(\omega_2)Q(\omega_2) + 1_A(\omega_3)Q(\omega_3) \\=1_A(\omega_1)X(\omega_1)P(\omega_1) + 1_A(\omega_2)X(\omega_2)P(\omega_2) + 1_A(\omega_3)X(\omega_3)P(\omega_3)\\=E_P(1_A X)  $$ where we used equivalence again to deal with the possibility that $P(\omega_i) = 0$ for some $i$. 
In the reverse direction, for any $X>0$, if there is an $A$ such that $Q(A) =0$ but $P(A)\ne 0,$ or vice versa then it's clear that $E_Q(1_A) \ne E_P(X 1_A)$ since one side is zero and the other isn't $(E_Q(1_A) = Q(A)$ and $E_P(X1_A) = 0$ if and only if $P(A) = 0$ since $X>0.$)
