There are $\binom{13}{3}$ ways to choose $3$ distinct ranks. Exactly $12$ of these will be straights (i.e. $A23\cdots QKA$) an must be excluded. There are $274$ distinct high-card hands in $3$-card poker.$$\binom{13}{3}-12=274$$ Another way to arrive at the same result, knowing that there are $16,440$ unique high-card hands, is to divide by the number of ways $(4^4-4=60)$ that such a hand is not a flush: $\frac{16440}{60}=274$. This is also the number of distinct flushes (rank of flush, disregarding suit): $\frac{1096}{4}=274$