Let $\overline{x y z y x}$ be the digits of the palindrome $a$. There are three cases. If $z<9$, then the next palindrome greater than $\overline{x y z y x}$ is $\overline{x y(z+1) y x}$, which differs by 100. If $z=9$ but $y<9$, then the next palindrome up is $\overline{x(y+1) 0}(y+1) x$, which differs from $\overline{x y 9 y x}$ by 110. Finally, if $y=z=9$, then the next palindrome after $\overline{x 999 x}$ is $\overline{(x+1) 000(x+1)}$, which gives a difference of 11. Thus, the possible differences are $11,100,110$.