Solution:

Count ordered triples $(a, b, c)$ with $abc = 11 \cdot 21 \cdot 31 \cdot 41 \cdot 51 \cdot 61 = 3^2 \cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61 = N.$ Each of the six primes $7, 11, 17, 31, 41, 61$ appears once and can go to any of the three values: $3^6$ ways. The two factors of $3$ can be split among the three values in $\binom{4}{2} = 6$ ways. That gives $6 \cdot 3^6 = 4374$ ordered triples.

If two of the three values were equal, their common value $v$ would satisfy $v^2 \mid N,$ so $v = 1$ or $v = 3.$ This produces the triples with values $\{1, 1, N\}$ and $\{3, 3, \frac{N}{9}\},$ each in $3$ orders, for $6$ ordered triples in all (all three equal is impossible). The remaining $4374 - 6 = 4368$ ordered triples have distinct entries, and each set $\{a, b, c\}$ is counted $3! = 6$ times.

So the number of sets is $\frac{4368}{6} = 728.$