Three. (20 points) Given
$$
(3 a+5 b-1)^{2}+|a+3 b+1|=0 \text {. }
$$

Find the solution set of the inequality with respect to $x$
$$
a x-b>\frac{x}{3}+6
$$
Three, from the conditions we know
$$
\left\{\begin{array} { l } 
{ 3 a + 5 b - 1 = 0 , } \\
{ a + 3 b + 1 = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=2, \\
b=-1 .
\end{array}\right.\right.
$$

Therefore, $2 x+1>\frac{x}{3}+6 \Rightarrow \frac{5}{3} x>5 \Rightarrow x>3$.
Thus, the solution set is $x>3$.