One way to approach this is to think adversarially: suppose that I'm the one who gets to start the three different trains of events, and I want you to wait as long as possible.
By renaming, let's assume that $t_a < t_b < t_c$. (The case where two or more are equal I leave to you). I'm going to say that you start waiting for your five events at time $T = 0$, but I'm going to be very particular here: If there's an event at time $0$, I want that to not count, i.e., you're looking for events in the interval $0 < T \le Q$ for some value of $Q$. Again, if you want to include the "starting moment", that's a case you have to work out for yourself.
By changing units of time, I'm going to assume $t_a$ is $1$. (I.e., if $t_a$ is 11 seconds, then I'm going to declare a gleep to be $11$ seconds, and say that $t_a$ is $1$ gleep, and measure $t_b$ and $t_c$ in gleeps as well.)
If $t_b$ and $t_c$ are huge (larger than $5$), you have to wait $5$ to observe $5$ events, for I, as your adversary, will put the first $A$ event at $T = 0$ (which you'll miss), another at $T = 1, 2, 3, 4, 5$, and by $T = 5$ you'll have observed one "A" event at each of those five times, so the total wait will be $5$. (or $5$ gleeps, if you like.)
But what if, as in your example, $t_b < 5$. Then I can avoid any $B$ events as long as possible by starting the $B$ sequence at $0$ as well --- you'll JUST miss the first one, notice a second one (the first that you count) at time $t_b$, another at time $2t_b$, and so on.
The same rationale applies to $t_c$: I can delay your seeing any $C$ events as long as possible by placing the first $C$ event at $T = 0$ as well.
What I'm doing in each case is trying to make sure that you have to wait as long as possible for five events to accumulate.
Let me simplify for a moment, and look at just TWO events, with $t_a = 1$ and $t_b = 2$. Drawing a picture, we have
*---*---*---*---*---*---*---
*-------*-------*-------*---
1 3 4 6 7
where the first row shows event $A$ happening every so often, and the second shows event $B$ happening twice as often, and the third row shows the count of events "since time 0" (ignoring the two events at time zero).
Notice here that there is never a moment when this count is exactly five.
The function that takes "time" to "events seen so far" is not 1-to-1, so there's not going to be a "formula" for the inverse. But there will be a good algorithmic way to compute the inverse, so it's OK.
Let's look at cases. (I apologize for this rambling approach...I'm developing the answer as I type).
Case 1: $t_b, t_c \ge 5$.
Solution: $T = 5$ is the longest you might have to wait for one event.
Case 2: $t_b < 5; t_c \ge 5$.
Solution: because $t_b < 5$, you're going to see at least one $B$ event before you get to the 5th $A$ event. If $t_b$ is between $4$ and $5$, then the $B$ event will be the fifth one you see, and we're done. If $t_b < 4$, then things are more subtle: you might see TWO $B$ events before time $5$. If the second one is between $4$ and $5$, then we'll ignore it, because we'll already have seen 4 $A$ events and 1 $B$ event. But if the second comes before $4$, then we have to count it.
So the solution for the case $t_c \ge 5, t_b < 5$ looks like this:
If $4 \le t_b < 5$, then you'll see (at least) 5 events in time $T = t_b$. Otherwise...
if $3 \le 2 t_b < 4$, then you'll see (at least) 5 events in time $T = 4$. Otherwise...
if $2 \le 2 t_b < 3$, you'll see 5 events in time $T = 2t_b$, Otherwise...
if $2 \le 3 t_b < 3$, then $T = 3$. Otherwise...
if $1 \le 3 t_b < 2$, then $T = 3t_b$ ...
I've probably got some of those cases wrong, but I'll bet you can work them out. The point is that after $S$ seconds, you'll have seen
$$
\newcommand{\floor}{\operatorname{floor}}
$$
- $\floor(S/t_a)$ events of type $A$
- $\floor(S/t_b)$ events of type $B$
- $\floor(S/t_c)$ events of type $C$
The sum of these is the number of events you'll have seen in $S$ units of time, when the adversary schedules the events as nastily as possible. You want to find the largest $S$ where this sum is five. The good news is that there are finitely many possibilities, all of them being multiples of $t_a, t_b, t_c$. The magic moment could be any of $t_a, 2t_a, 3t_a, \ldots, 5t_a$, $t_b, 2t_b, \ldots, 5t_b$, $t_c, \ldots, 5t_c$. That's exactly 15 cases to check. You can slightly reduce this by letting
$$
M = \floor(5 t_a / t_b) \\
K = \floor(5 t_A / t_c).
$$
Then you need to check
$$
t_a, 2t_a, \ldots, 5t_a\\
t_b, 2t_b, \ldots, M t_b \\
t_c, 2t_c, \ldots, Nt_c
$$
where each of $M$ and $N$ will be at most 5, but probably less (and could be less than $1$, in the case where $t_b$ and $t_c$ are "large" (i.e., greater than $5 t_a$).
Summary: there's not a formula, but there's a very well-bounded computational problem to solve. If you wanted to know how long you had to wait for $3187$ events, it'd be a mess. For five? Not so bad.
