To find all connected 6-fold cover of $\mathbb R P^2 \# \mathbb R P^2 \# \mathbb R P^2 $ What is the full list of connected 6-fold covers of $\mathbb R P^2 \# \mathbb R P^2 \# \mathbb R P^2$, up to covering isomorphism?
This is equivalent to finding all subgroups $H$ of $G = \langle a,\ b,\ c \ | \ aabbcc = 1 \rangle$ such that $[G:H]=6$.
Letting $P := \mathbb RP^2$ and $T$ the torus, then $P \# P \# P \simeq T \# P$, so $$ T\# P\# P\# P\# P\# P\# P \ \ \ \text{or} \ \ \ T \# T \# T$$are possible ones, I think. But how to get the full list? Topological way would be good, and algebraic either.
 A: Your last paragraph has an implied question which differs from the question asked in your first paragraph. That implied question is: What are all the actual covering spaces... up to abstract homeomorphism? The answer you propose is half right: there are two of them, but you listed one of them incorrectly.
The surface $P \# P \# P$ is a closed, connected, non-orientable surface of Euler characteristic $-1$. So a connected 6-fold covering space will be a closed, connected surface, either orientable or non-orientable, of Euler characteristic $-6$. According to the classification of surfaces there are two possibilities, up to abstract homeomorphism: one orientable, homeomorphic to a connected sum of four $T$'s, $T \# T \# T \# T$ (not three, as your answer has it); or a connected sum of three $T$'s and two $P$'2, $T \# T \# T \# P \# P$ which, using the identities you mentioned, is homeomorphic to a connected sum of one $T$ and six $P$'s (as your answer has it) and also, incidentally, homeomorphic to a connected sum of eight $P$'s.
But this does not yet address the question in your first paragraph: What are all of the 6-fold coverings up to covering isomorphism? This requires constructing all of the different covering maps $T \# T \# T \# T \mapsto P \# P \# P$ up to covering equivalence, and similarly for $T \# T \# T \# P \# P \mapsto P \# P \# P$. There is more than one of each!!
For this, let me just get you off the ground. Every non-orientable surface has a unique orientable degree 2 covering map up to covering equivalence. The degree 2 covering space of $P \# P \# P$ is the closed, connected, orientable surface of Euler characteristic $-2$ which is $T \# T$. Every orientable covering map over $P \# P \# P$ must factor uniquely through this orientable degree two covering map, and so for the orientable case you have reduced your problem to the construction of covering maps $T \# T \# T \# T \to T \# T$ up to covering equivalence. There is still more than one of these!! To find them all, it's time to start doing some algebra, and then you have to do it all again for constructing all of the covering maps $P \# P \# P \# P \# P \# P \# P \# P \mapsto P \# P \# P$. But I'll stop here.
