Since $360=2^{3} \cdot 3^{2} \cdot 5$, the only possible prime divisors of $d_{i}$ are 2,3 , and 5 , so we can write $d_{i}=2^{a_{i}} \cdot 3^{b_{i}} \cdot 5^{c_{i}}$, for nonnegative integers $a_{i}, b_{i}$, and $c_{i}$. Then, $d_{1} d_{2} d_{3} \mid 360$ if and only if the following three inequalities hold. $$\begin{aligned} a_{1}+a_{2}+a_{3} & \leq 3 \\ b_{1}+b_{2}+b_{3} & \leq 2 \\ c_{1}+c_{2}+c_{3} & \leq 1 \end{aligned}$$ Now, one can count that there are 20 assignments of $a_{i}$ that satisfy the first inequality, 10 assignments of $b_{i}$ that satisfy the second inequality, and 4 assignments of $c_{i}$ that satisfy the third inequality, for a total of 800 ordered triples $(d_{1}, d_{2}, d_{3})$.