jmc

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, $$\begin{array}{rl}(a+1{)}^2+(6a{)}^2& =(6a+1{)}^2⟹\\ a^2+2a+1+36a^2& =36a^2+12a+1⟹\\ a^2+2a+\cancel{1}+\cancel{36a^2}& =\cancel{36a^2}+12a+\cancel{1}⟹\\ a^2-10a& =0⟹\\ a(a-10)& =0⟹\\ a=0& \text{or}a=10.\end{array}$$ Take the positive solution $a=\boxed{10}$.