Find the number of triplets of integers in arithmetic progression, the sum of wh

Let the three integers in arithmetic progression be a-d, a, a+d. Their sum of squares is 3a^2 + 2d^2 = 1994. Since 2d^2 is even, 3a^2 must be even so a is even. Let a=2k, then 12k^2+2d^2=1994 => 6k^2+d^2=997. For k=0 to 12, d^2 is never a perfect square. Hence no integer solutions exist.