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If all lines satisfy the condition, then we can just plug in values for $a$, $b$, and $c$ that form an arithmetic progression. Let's use $a=1$, $b=2$, $c=3$, and $a=1$, $b=3$, $c=5$. Then the two lines we get are: $$x+2y=3$$ $$x+3y=5$$ Use elimination to deduce $$y=2$$ and plug this into one of the previous line equations. We get $$x+4=3\Rightarrow x=-1$$ Thus the common point is $\boxed{(-1,2)}$.