Reasoned solution:

Let the blanks be filled in with $a_0$, $a_1$, $a_2$, $a_3$, $a_{\text{ev}}$, $a_{\text{od}}$, and $a_{\text{pr}}$, respectively. Among these, the first four entries satisfy $a_i\ge 1$ for $i\le 3$, while the other three entries each satisfy $a\ge 2$. Thus, $a_0=1$, $a_1\ge 2$, and $a_{\text{od}}\ge 3$. Given $a_{\text{ev}}+a_{\text{od}}=11$, the values have opposite parity, so there’s at least one additional even and odd entry. Consequently, $a_{\text{ev}}\ge 3$ and $a_{\text{od}}\ge 4$. This limits $(a_{\text{ev}},a_{\text{od}})$ to only five pairs: $(3,8),(4,7),(5,6),(6,5),$ or $(7,4)$. Each pair has exactly one prime number, hence $a_{\text{pr}}\ge 3$, but $a_{\text{pr}}\ne 3$, as this “3” would add a fourth prime to the set, implying $a_{\text{pr}}\ge 4$.

Now, only $a_2$ and $a_3$ are potential placements for “1,” so $a_1\le 4$. If $a_1=4$, there are three primes, including one from $(a_{\text{ev}},a_{\text{od}})$, but not $a_{\text{pr}}$. Setting $a_{\text{pr}}=4$ would then leave only three primes, and any other $a_{\text{pr}}$ would be too large. Thus, $a_1\le 3$. Assume $a_1=3$; then $a_3\ge 2$. To assess feasibility, suppose $a_3=2$, which implies $a_2\ge 2$, creating a contradiction with $a_1=3$. Thus, $a_3\ge 3$. If $a_3\ge 4$, $a_{\text{pr}}=3$ leads to more than three primes, so $a_3=3$.

With $a_3=3$, counting primes from $(a_{\text{ev}},a_{\text{od}})$ yields five primes. Setting $a_{\text{pr}}=5$ results in six primes, while $a_{\text{pr}}=6$ yields five primes, a contradiction. Therefore, $a_1=2$, implying $a_2\ge 3$ because $a_2=2$ would introduce three “2”s. Since only $a_3=2$ allows a third 2, we conclude $a_2=3$ and $a_3=2$.

Thus, we find six primes with $a_{\text{pr}}=6$ or $a_{\text{pr}}=7$, resulting in two possible solutions: $(1,2,3,2,6,5,6)$ and $(1,2,3,2,5,6,7)$.