Can you prove why the recurrence $3P_{n} - 2P_{n + 1} = P_{n - 1}$ holds for some (a subset of the) prime numbers?

Using the standard prime gap notation gives that

$$g_n = p_{n+1} - p_{n} implies p_{n+1} = p_{n} + g_{n} tag{1}label{eq1A}$$

$$p_{n} = p_{n-1} + g_{n-1} tag{2}label{eq2A}$$

Next, eqref{eq2A} in eqref{eq1A} gives

$$p_{n+1} = p_{n-1} + g_{n-1} + g_{n} tag{3}label{eq3A}$$

Substituting eqref{eq2A} and eqref{eq3A} into your relation results in

$$begin{equation}begin{aligned} & 3p_{n} - 2p_{n + 1} = p_{n - 1} \ & 3(p_{n-1} + g_{n-1}) - 2(p_{n-1} + g_{n-1} + g_{n}) = p_{n - 1} \ & 3p_{n-1} + 3g_{n-1} - 2p_{n-1} - 2g_{n-1} - 2g_{n} = p_{n - 1} \ & p_{n-1} + g_{n-1} - 2g_{n} = p_{n - 1} \ & g_{n-1} = 2g_{n} end{aligned}end{equation}tag{4}label{eq4A}$$

Thus, your relation holds whenever any prime gap is double that of the next prime gap. With your example of $n = 5$, we have $g_{n-1} = 11 - 7 = 4$ and $g_{n} = 13 - 11 = 2$.

As for whether or not there's an infinite number of primes where eqref{eq4A} holds, I'm quite certain that is currently unknown, but it's conjectured to be the case. Note there are infinitely many prime $k$-tuples which are admissible and, in the cases where they also represent a prime constellation, then eqref{eq4A} is true, e.g., your example of $(0, 4, 6)$ (which is called a prime triplet), $(0, 8, 12)$, $(0, 12, 18)$, etc. The "Admissibility" section also states

It is conjectured that every admissible $k$-tuple matches infinitely many positions in the sequence of prime numbers. However, there is no admissible tuple for which this has been proven except the $1$-tuple ($0$). Nevertheless, by Yitang Zhang's famous proof of $2013$ it follows that there exists at least one $2$-tuple which matches infinitely many positions; subsequent work showed that some $2$-tuple exists with values differing by $246$ or less that matches infinitely many positions.^[2]

In particular, the prime $k$-tuple conjecture states

... that every admissible pattern for a prime constellation occurs infinitely often ...

This conjecture implies there are an infinite number of primes which satisfy eqref{eq4A}.