It is given that $\log_{6}a + \log_{6}b + \log_{6}c = 6,$ where $a,$ $b,$ and $c$ are [positive](https://artofproblemsolving.com/wiki/index.php/Positive) [integers](https://artofproblemsolving.com/wiki/index.php/Integer) that form an increasing [geometric sequence](https://artofproblemsolving.com/wiki/index.php/Geometric_sequence) and $b - a$ is the [square](https://artofproblemsolving.com/wiki/index.php/Perfect_square) of an integer. Find $a + b + c.$
$abc=6^6$. Since they form an increasing geometric sequence, $b$ is the [geometric mean](https://artofproblemsolving.com/wiki/index.php/Geometric_mean) of the [product](https://artofproblemsolving.com/wiki/index.php/Product) $abc$. $b=\sqrt[3]{abc}=6^2=36$.
Since $b-a$ is the square of an integer, we can find a few values of $a$ that work: $11, 20, 27, 32,$ and $35$. Out of these, the only value of $a$ that works is $a=27$, from which we can deduce that $c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48$.
Thus, $a+b+c=27+36+48=\boxed{111}$