One can choose 3 objects out of 28 objects in (28,3) ways.
Among these choices all would be together in 28 cases; exactly two will be together in 28 x 24 cases.
Thus three objects can be chosen such that no two adjacent in (28,3) -28 - (28 x 24) ways.
Among these, furthrer, two objects will be diametrically opposite in 14 ways and the third would be on either semicircle in a non adjacent portion in 28 - 6 = 22 ways.
Thus required number is
(28,3) - 28 - (28 x 24) - (14 x 22) = 2268: