Answer: 331

Solution:

The ordered triples $(a, b, c)$ of positive integers with $a + b + c = 66$ number $\binom{65}{2} = 2080.$ Exactly one of them has all three values equal, namely $(22, 22, 22).$ Triples with exactly two values equal come from $2a + c = 66$ with $c \ne a:$ here $a$ can be $1$ through $32$ except $22,$ giving $31$ multisets, each arrangeable in $3$ ways, so $93$ ordered triples.

Hence $2080 - 1 - 93 = 1986$ ordered triples have three distinct values, and each unordered choice $a \lt b \lt c$ is counted $6$ times. The number of valid separations is $\frac{1986}{6} = 331.$