First, let me shorten their names by referring to them by their first letter, $T$, $B$ and $S$.
Let us solve this problem using states.
For those of you who have not come across this concept before, I have posted a link below in the comments which is the wiki from which I learnt this problem solving method. I really recommend it as it has been very helpful for me!
Nevertheless, I'll try to explain the concept in this spoiler for those of you who are new to it.
In this game, there are different 'states'. For example, it might be $T$ to roll. For each of these different states, I work out the probability that an event occurs in that state, like $T$ getting the next 6. Often, I will write this in terms of $T$ getting a 6 in other states. Eventually, I can combine the equations to give me a numerical answer.
States are especially useful in cases when you have no idea when the game will end (i.e. in games where it is a constant loop where there is some chance of the game ending on each round).
Let $E(X_Y)$ denote the probability $X$ will roll the next 6 when $Y$ is the next person the roll the die.
Let us first consider the probability that $T$ will roll the the first 6:
The game starts with the state where $T$ is about to roll, and we are interested in the chances that $T$ will roll the next 6, so we want to find $E(T_T)$.
Using the above 3 equations:
We are only interested in what happens when $T$ rolls the first 6. Any other cases are not relevant.
So now we need to find the probability that $B$ rolls a 6 given that $T$ has already rolled a 6. The game is now reduced to 2 players with it being $B$'s turn (as $T$ just rolled a 6).
Note that this is a different game from before as it has changed from 3 to 2 players, so $E(X_Y)$, in general, has a different value now, since we assume $T$ is out. For example, $E(T_B)=0$ now, since $T$ cannot roll a 6, of course (as $T$ is out of the game).
Now we have the new equations:
Now if $T$ gets the first 6, and then $B$ gets the next 6, then $S$ cannot fail to get the 6 after that ($S$ will eventually roll one, even if it takes over 9000 tries).
So therefore, the probability that order will happen is: