If $$log_{25}{5}$$ = a and $$log_{25}{15} $$ = b, then the value of $$log_{25}{27}$$ is:

$$log_{25}{5}$$ = a 

=> a=1/2

$$log_{25}{15}$$ = $$log_{25}{3}+log_{25}{5}$$ = b

$$\frac{1}{2} \log_{5}{3} + \frac{1}{2}$$ = b

$$\log_{5}{3}$$ = 2(b -  $$\frac{1}{2})$$.............(i)

$$log_{25}{27}$$ = $$\frac{3}{2} \log_{5}{3}$$.........(ii)

Replacing $$\log_{5}{3}$$ = 2(b -  $$\frac{1}{2})$$ in (ii) we get

$$log_{25}{27}$$ = 3(b -  $$\frac{1}{2}$$)

We can write -$$\frac{1}{2}$$ as (- 1 + $$\frac{1}{2}$$) or (-1 + a)

So, $$log_{25}{27}$$ = 3(b + a - 1)

Hence, option C is the correct answer.