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Yes, you can do this in awk:

awk '{ k=0; for(i=2;i<=NF;i++){ if($i == 0){ k++ } } if(k==3){ tot++ } } END{ print tot }' file 

But, personally, I would do in in perl instead:

$ perl -ale '$tot++ if (grep{$_ == 0 } @F) == 3 }{ print $tot' file 7 

Or, the slightly less condensed:

$ perl -ale 'if( (grep{$_ == 0 } @F) == 3 ){ $tot++ } END{ print $tot }' file 7 

And the same thing, for the golfers among you:

$ perl -ale '(grep{$_==0}@F)==3&&$t++}{print$t' file 7 

Explanation

• -ale: -a makes perl behave like awk. It will read each line of the input file and split it on whitespace into the array @F. The -l adds a \n to each call of print and removes trailing newlines from the input and the -e is the script that should be applied to each line of input.
• $tot++ if (grep{$_ == 0 } @F) == 3 : increment $tot by one, for every time where there are exactly 3 fields that are 0. Since the 1st field starts from 1, we know it will never be 0 so we don't need to exclude it.
• }{: this is just a shorthand way of writing END{}, of giving a block of code that will be executed after the file has been processed. So, }{ print $tot will print the total number of lines with exactly three fields with a value of 0.