1. Positive integers $a, b, c$ satisfy
$$
\log _{6} a+\log _{6} b+\log _{6} c=6 \text {, }
$$
$a, b, c$ form an increasing geometric sequence, and $b-a$ is a perfect square. Then the value of $a+b+c$ is $\qquad$
$=.1 .111$.
From the given, we have $abc=6$, and $b^2=ac$. Therefore, $b=36$.
Also, $b-a$ is a perfect square, so $a$ can take the values $11, 20, 27, 32$.
Upon verification, only $a=27$ meets the conditions. In this case, $c=48$.
Thus, $a+b+c=27+36+48=111$.