Probability of data loss for mirrored,3 disk system
Let us look at the probability of the "123" event happening. That is, Disk 1 fails, and then Disk 2, and then Disk 3, and all of these happen within a span on $H$ hours. Your claim is that the probability of this is $F ∗ (\frac{FH}{365∗24})^2$. But the $\frac{FH}{365∗24}$ term only denotes the probability of a disk failing in the $H$ hours after the first  disk fails. In particular, it does not consider the ordering constraint of "Disk 2 fails before Disk 3 fails". 

Instead, what $F ∗ (\frac{FH}{365∗24})^2$ denotes is the probability that Disk 1 fails, and in the ensuing $H$ hours, both Disk 2 and Disk 3 fail. So this captures both the events "123" and "132".

Similarly, the probability of either the event "213" or "231" happening would be $F ∗ (\frac{FH}{365∗24})^2$. And similarly for the events "312" and "321" combined.

So the final answer should be $(F ∗ (\frac{FH}{365∗24})^2) * 3$, and not $(F ∗ (\frac{FH}{365∗24})^2) * 6$.