What are the quiescent states of the surface code? Notation
$$X \equiv \sigma_x \qquad Z \equiv \sigma_z$$
Introduction
The surface code is a quantum error correction code.
In the surface code, we have a two-dimensional grid of qubits each coupled to its four nearest neighbors.
As shown in Fig. 1, within a row or column, every other qubit is alternatingly a data or measure qubit.
On each round of the surface code's error correction mechanism, the measure qubits each measure the 4-way parity of either $X$ or $Z$ of their four neighboring data qubits.
The essence of the surface code is contained in the fact that these measurements do not completely collapse the state of the grid.
This can be seen in a 2-qubit example.
The two measure qubits highlighted in Fig 1 measure the two operators
$$X_A X_B \quad \text{and} \quad Z_A Z_B \, . $$
These operators commute, so there are in fact simultaneous two-qubit eigenstates of those parity operators.
In fact, those eigenstates are precisely the four two-qubit Bell states.
Question
We know that in the the two qubit case, there are four possible eigenstates of the parity measurements, i.e. the Bell states.
What are the eigenstates with more than two qubits?
As an example, we could consider a small patch of the surface code that looks like this:
X o X
o Z o

where o means a data qubit, X means a measure-X qubit, and Z means a measure-Z qubit.
Another "simple" case could be
o X o X
Z o Z o
o X o X
Z o Z o

with periodic boundary conditions.


Figure 1
 A: Let's take your example:
X o X
o Z o

So three data qubits (the o's). Let's label them 0, 1, 2, left to right. This little surface code involves measuring the operators $X_0X_1$, $Z_0Z_1Z_2$, $X_1X_2$. Working out explicit states is straightforward. Start with the all 0 state, as this is a simultaneous +1 eigenstate of all Z parity operators:
$$|000 \rangle \, .$$
Then systematically build a state that is also the +1 eigenstate of all X parity operators. First $X_0X_1$:
$$\frac{1}{\sqrt{2}}(|000\rangle + X_0 X_1|000\rangle) = \frac{1}{\sqrt{2}}(|000\rangle + |110\rangle)$$
Then $X_1X_2$:
$$\frac{1}{2}[(|000\rangle + |110\rangle) + X_1X_2(|000\rangle + |110\rangle)] = \frac{1}{2}(|000\rangle + |110\rangle + |011\rangle + |101\rangle) \, .$$
To obtain the state that is the -1 eigenstate of any particular parity operator, find a path of Pauli operators that anticommutes only with it, and apply these operators to the state. For example, $Z_0$ anticommutes only with $X_0X_1$, so the simultaneous -1 eigenstate of $X_0X_1$ and +1 eigenstate of both $Z_0Z_1Z_2$ and $X_1X_2$ is:
$$\frac{1}{2}(|000\rangle - |110\rangle + |011\rangle - |101\rangle) \, .$$
At least one such path exists for every parity operator, and it doesn't matter which path you choose, you'll always get the same state.
