Calculating probability of reaching state in DTMC
The answer to your question very much depends on the values of the probabilities $a_{ij}$. If all are non-zero, then the DTMC forms one single strongly connected component and the probability of reaching $x=3$ is therefore $1$ (it is $1$ for any state in such a case in fact). This result comes from the Long-run theorem, which says that any finite DTMC eventually reaches a bottom SCC and visits all its states infinitely often.

Have a look at the lecture slides and notes of a course on probabilistic systems. Especially the third and fourth lecture.

In a general case, you would find with a graph analysis states that reach $x=3$ with probability $1$ denoted $S_1$ and those with probability $0$ denoted $S_0$. Then you have a set of states $S_?$ for which you don't yet know the probability. For them you construct a matrix $\mathbf{A}$ of transition probabilities (it will be a square matrix of size $|S_?|\times |S_?|$). Then you construct a vector $\mathbf{b}$ with one-step probability to reach $x=3$. A solution to $\mathbf{A}\cdot \mathbf{x} = \mathbf{b}$ gives you the probabilities for states in $S_?$.