Mark is correct. There is no LOS to H6, but LOS to the other two. The calculation goes as this:
I6 (the obstacle in front of H6) creates one blind hex since the distance from O5 is 5-9 hexes (B10.23). O5 is exactly one level higher than I6, and will therefore not reduce the number of blind hexes - thus no LOS.
G7 (the obstacle in front of F7) also creates one blind hex, but according to A6.42: take the height differences between O6 and G7 (4-2 = 2), subtract by one (2-1 = 1), and reduce the number of blind hexes by the last number (1-1 = 0), thus no blind hexes.
Similarily for D6, where the obstacle is E6. The number of blind hexes is 2 (distance 10), but height difference of O6 and E6 is 3. 3 - 1 = 2, so reduce the number of blind hexes by 2 - again zero blind hexes.