If $$log(2^{a}\times3^{b}\times5^{c} )$$is the arithmetic mean of $$log ( 2^{2}\times3^{3}\times5)$$, $$log(2^{6}\times3\times5^{7} )$$, and $$log(2 \times3^{2}\times5^{4} )$$, then a equals

$$log(2^{a}\times3^{b}\times5^{c} )$$ = $$ \frac{log ( 2^{2}\times3^{3}\times5) + log(2^{6}\times3\times5^{7} ) + log(2 \times3^{2}\times5^{4} ) }{3} $$

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

$$log(2^{a}\times3^{b}\times5^{c} )$$ = $$ \frac{log ( 2^{9}\times3^{6}\times5^{12}) }{3} $$

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