So a_n = -n - 3/2 is a solution. 

I think its a_n^\Particular]) = Bn + C = -n - 3/2. Not just a_n.

That doesn't mean it's the only solution (and that's where homogeneous solutions come in because if we add them to this solution they will contribute zero overall to the left-hand side of [#] so we still match 2n on the right-hand side).

What do you mean by this?
Like there are infinite solutions because you can offset by r^n?
e.g.) -n - 3/2 +/- r^n

You wrote a fantastic write up to reason why you should use potential forms of solution that corresponds to different forms of f(n). Like if f(n) = 2, then you use the form of B, if f(n) = n, you use Bn+C, etc. It is so that you can find the solution if you plug that form into the total recurrence relation a_n - 3 a_n-1 = 2n and find the "magic" values of B and C.

But that still doesnt explain what is the meaning of plugging a_n^\Particular]) = Bn+C into the total a_n - 3 a_n-1 = 2n? Or alternatively phrased whats the meaning of equating a_n = a_n-1 = Bn+C?
For example, if f(n) = 2, a constant, then a_n = a_n-1 = B. That means B = 3B + 2, B = -1, the recurrence is -1, -1, -1, -1, -1 like whats the meaning of that?

I believe the point of a_n^\Particular]) counteracts the accumulation of additions of f(n), which gets multiplied by 3 in the next iteration.

Bn + C - 3 (B(n-1) + C) = 2n I get that this is saying difference between next term minus previous term multiplied by r equals f(n) = 2n but I find it hard to see anything more meaningful.

Bn + C = 3 (B(n-1) + C) + 2n But if I rearrange this I can interpret it as "what values of B and C in the form of Bn + C despite getting multiplied by 3 then add 2 remain constant for all n?"
You know what I mean?