You have carried out your example without using the system you described in the article you linked.
Ballots:
66 A > B > C
34 B > C > A
Borda scores:
A: 132
B: 134
C: 34
Under Borda, candidate B wins.
Your method:
Ignore {}: (A=66, B=34, C=0, demotions=0), no one reaches quota
Ignore {A}: (A=0, B=100, C=0, demotions=66), B reaches quota
Ignore {B}: (A=66, B=0, C=34, demotions=34), no one reaches quota
Ignore {C}: (A=66, B=34, C=0, demotions=0), no one reaches quota
Ignore {A,B}: (A=0, B=0, C=100, demotion=166), C reaches quota
Ignore {A,C}: (A=0, B=100, C=0, demotions=66), B reaches quota
Ignore {B,C}: (A=100, B=0, C=0, demotions=68), A reaches quota
The are four possible sets-to-ignore which result in a candidate reaching quota:
Ignore {A}: B wins after 66 demotions
Ignore {A,B}: C wins after 166 demotions
Ignore {A,C}: B wins after 66 demotions
Ignore {B,C}: A wins after 68 demotions
We identify from those sets the ones which minimize choice demotions:
Ignore {A}: B wins after 66 demotions
Ignore {A,C}: B wins after 66 demotions
Both cause B to win, and per your tie-break procedure we prefer the set which affects the fewest ballots. Both sets affect 66 ballots. The second tiebreaker is “most candidates”, which means our chosen set is {A,C}.
Per your own written and published explanation of your voting system, the outcome of this election is for B to win when we ignore {A,C}, with 66 choice demotions.
As I said before, there is no room for debate about the fact that your system is exactly equivalent to the Borda count (in the single-winner case where all voters rank all candidates), because the procedure by which you defined your system necessarily yields the Borda count when the number of choice demotions is subtracted from a constant (in this case 200, in general (n-1)*v).
I strongly recommend that you move away from your repeatedly-demonstrated propensity to argue against mathematical facts.
Nevin