Since $ 6a<6a+1 $ for all real numbers $ a $, the length of the hypotenuse cannot be $ 6a $ units. Also, $ a $ is positive because one of the side lengths is $ 6a $ units. Since $ a+1 < 6a+1 $ for all positive real numbers $ a $, the length of the hypotenuse cannot be $ a+1 $ either. Therefore, $ a+1 $ and $ 6a $ are the legs of the right triangle, and by the Pythagorean theorem, \begin{align*} (a+1)^2+(6a)^2&=(6a+1)^2 \implies \\ a^2+2a+1+36a^2&=36a^2+12a+1 \implies \\ a^2+2a+\cancel{1}+\cancel{36a^2}&=\cancel{36a^2}+12a+\cancel{1} \implies \\ a^2-10a&=0 \implies \\ a(a-10)&=0 \implies \\ a=0\qquad&\text{or}\qquad a=10.\end{align*} Take the positive solution $ a=\boxed{10} $.