Alkyne formation from vicinal dihalide
I think the comments basically have the answer - if you use only 2 equivalents of sodium amide, you will use 1.33 equivalents creating the alkyne and the remaining 0.67 equivalent will react with the alkyne after it is formed (an acid-base reaction) to form the acetylide salt -- probably a much faster reaction than the dehydrobromination reactions. So about 1/3 of the original dibromide will remain unreacted. You need the third equivalent of sodium amide to drive the reaction to completion.