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 \geq 1 for i \leq 3, while the other three entries each satisfy a \geq 2. Thus, a_0 = 1, a_1 \geq 2, and a_{\text{od}} \geq 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}} \geq 3 and a_{\text{od}} \geq 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}} \geq 3, but a_{\text{pr}} \neq 3, as this “3” would add a fourth prime to the set, implying a_{\text{pr}} \geq 4.

Now, only a_2 and a_3 are potential placements for “1,” so a_1 \leq 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 \leq 3. Assume a_1 = 3; then a_3 \geq 2. To assess feasibility, suppose a_3 = 2, which implies a_2 \geq 2, creating a contradiction with a_1 = 3. Thus, a_3 \geq 3. If a_3 \geq 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 \geq 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).