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$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{2}\times3^{3}\times5) + log(2^{6}\times3\times5^{7} ) + log(2 \times3^{2}\times5^{4} ) }{3} $
$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{2+6+1}\times3^{3+1+2}\times5^{1+7+4}) }{3} $
$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{9}\times3^{6}\times5^{12}) }{3} $
$3log(2^{a}\times3^{b}\times5^{c} )$ = $ log ( 2^{9}\times3^{6}\times5^{12}) $Hence, 3a = 9 or a = 3