Answer: 70

Small Hint:

Treat the exponents of $2$ and $5$ independently

Big Hint:

For each prime, translate every least common multiple into a condition on pairwise maxima

Solution:

For the exponent of $5,$ all three pairwise maxima equal $3.$ Thus at least two exponents are $3:$ there is one all-$3$ triple and $3\cdot3$ triples with exactly two $3$’s, for $10$ choices.

For the exponent of $2,$ the first pair has maximum $3$ while the other two have maximum $4.$ The exponent of $c$ must be $4,$ and the exponents of $a$ and $b$ lie in $\{0,1,2,3\}$ with maximum $3,$ giving $4^2-3^2=7$ choices. Independence gives $10\cdot7=70.$