jmc

By the properties of logarithms, $${\log }_{6}a+{\log }_{6}b+{\log }_{6}c={\log }_6(abc)=6,$$so $abc=6^6.$ But $(a,b,c)$ is an increasing geometric sequence, so $ac=b^2,$ and $abc=b^3=6^6.$ Thus, $b=6^2=36.$

Therefore $b-a=36-a$ is a nonzero perfect square. We also have $c=b^2/a=6^4/a,$ so $a$ must be a divisor of $6^4.$ Testing perfect square values for $36-a,$ we find that the only possible value of $a$ is $a=27,$ giving $c=6^4/27=48.$ Thus, $$a+b+c=27+36+48=\boxed{111}.$$