11. The smallest prime \( p \) (where \( p > 3 \)) for which there do not exist non-negative integers \( a, b \) satisfying \( \left|3^{a}-2^{b}\right|=p \) is
11. 41

Hint: $2=\left|3^{1}-2^{0}\right|, 3=\left|3^{0}-2^{2}\right|$,
$$
\begin{array}{l}
5=\left|3^{2}-2^{2}\right|, 7=\left|3^{2}-2^{1}\right|, \\
11=\left|3^{3}-2^{4}\right|, 13=\left|3^{1}-2^{4}\right|, \\
17=\left|3^{4}-2^{6}\right|, 19=\left|3^{3}-2^{3}\right|, \\
23=\left|3^{3}-2^{2}\right|, 29=\left|3^{1}-2^{5}\right|, \\
31=\left|3^{0}-2^{5}\right|, 37=\left|3^{3}-2^{6}\right| .
\end{array}
$$

It remains to prove that there do not exist $m, n \in \mathbf{N}$, such that $41=\left|3^{m}-2^{n}\right|$ (consider two cases).