Question 457998
From the given, {{{2^a = 12^c}}} ==> a*ln2 = c*ln12 ==> {{{c = (aln2)/ln12}}}.

Also, from {{{3^b = 12^c}}} ==> b*ln3 = c*ln12 ==> {{{b*ln3 = ((aln2)/ln12)*ln12}}} ==> {{{b = (aln2)/ln3}}}

Then {{{1/c - 1/b - 2/a = 1/((aln2)/ln12) - 1/((aln2)/ln3 ) - 2/a = (1/a)*(ln12/ln2 - ln3/ln2 - 2) = (1/a)*(ln4/ln2 - 2) = (1/a)*(2-2) = 0}}}