Question about proper subgames in game theory. Consider the following game presented as a tree.
As we can see there are $3$ players and the question is to find all N.E and all S.P.N.E. But I have a doubt about one thing. As we can see, if the first player selects $L$ then there are no actions for the second player. Am I right that in this game there is no $(L,L,L)$ strategy? Or if there is such strategy, then it's utility is (3,3,2)? If the last variant is valid, then how can we produce any proper subgame for this tree?

 A: Strategies in seqential games of perfect information are phrased a bit differently than in a simultaneous move game - after all, actions of the players moving later can condition on what happened before.
Hence, the stategies of the third player in such a sequential game are more appropriately described as follows: $L$ if 1 chose $L$, $R$ if 1 chose $R$ and 2 chose $L$.
Similarly, the strategy of the second player would then be described as: $L$ if 1 chose $R$. This makes clear that player 2 does not get to move if 1 chose $L$.
So a strategy profile in this sequential game would not be $(L,L,L)$, nor even $(L,L)$, but indeed $(L, L\textrm{ if ... and } R \textrm{ if ...}, L \textrm{ if ... and } R \textrm{ if ...})$ (for example).
This conditional phrasing of strategies is important when describing Nash equilibrium. It needs to be clear what would happen in another branch of the tree if that were chosen, hence strategies in the off-equilibrium paths need to be specified as well, even if the game never ends up there in equilibrium. But it is important to establish that going there is not a profitable deviation, i.e., is important to establish the equilibrium.
For example, $(L, R \textrm{ if 1 chose }R, L \textrm{ if 1 chose }L \textrm{ and anything otherwise})$ is a Nash equilibrium. That it is a NE depends very much on the strategy of player 2 who does not even get to move on the equilibrium path in this equilibrium, because the outcome is $L$ for 1 and $L$ for 3. So if 2's strategy was $L$ instead, then 1 might profitably deviate from $L$ to $R$, which shows that even players who do not get to move on the equilibrium path are still important for the equilibrium, because all players react to each other.
