The first part is correct, but you need to exclude the possibility of having a second triple within the other four cards.
C(exactly one triple) = C(triple) - C(pair triples)
C(triple)
Choose a rank for "triple": C(13,1)
Choose three cards for "triple": C(4,3)
Choose four cards not "triple": C(48,4)
C(pair triples)
Choose a rank for "each" triple: C(13,2)
Choose three cards for "first" triple: C(4,3)
Choose three cards for "second" triple: C(4,3)
Choose one card not from "pair triples": C(44,1)
*52 cards in deck; first triple = 4 cards, second triple = 4 cards, remaining cards to choose that are different from the pair of triples: 52-4-4=44
C(triple) = C(13,1)*C(4,3)*C(48,4) = 13 x 778,320
C(pair triples) = C(13,2)*C(4,3)^2*C(44,1) = 78 x 704
C(exactly one triple) = 10,118,160 − 54,912 = 10,063,248 ways
If you want to know the probability: just divide the above, by the number of ways to choose 7 cards from a deck of 52 = C(52,7) = 133,784,560
P(exactly one triple) = Actual Outcomes / Total Outcomes = 10,063,248 / 133,784,560 = 0.075219.. ~7.5%