We are given three languages over the alphabet $ \{0,1\} $, with a delimiter symbol $ \# \notin \{0,1\} $:
- $ L_1 = \{ w w^R \mid w \in \{0,1\}^* \} $
- $ L_2 = \{ w \# w^R \mid w \in \{0,1\}^* \} $
- $ L_3 = \{ w w \mid w \in \{0,1\}^* \} $
Language $ L_1 $:
$ L_1 $ is the set of even-length palindromes. It is context-free, as shown by the grammar
$$
S \to 0S0 \mid 1S1 \mid \varepsilon.
$$
However, it is not deterministic. A deterministic pushdown automaton (DPDA) cannot identify the middle of the input without a marker and must guess when to switch from pushing to popping. Since this guess cannot be made deterministically, $ L_1 \notin \text{DCFL} $. Thus,
$$
L_1 \in \text{CFL} \setminus \text{DCFL}.
$$
Language $ L_2 $:
The symbol $ \# $ marks the center of the string. A DPDA can deterministically push symbols onto the stack while reading $ w $, and upon reading $ \# $, switch to popping mode to match $ w^R $. This yields a DPDA for $ L_2 $, so $ L_2 \in \text{DCFL} $. A context-free grammar is:
$$
S \to 0S0 \mid 1S1 \mid \#.
$$
Language $ L_3 $:
Assume $ L_3 $ is context-free. Let $ p $ be the pumping length. Consider
$$
s = 0^p 1 0^p 1 \in L_3,
$$
where $ w = 0^p 1 $. By the pumping lemma for context-free languages, $ s = uvxyz $ with $ |vxy| \leq p $, $ |vy| \geq 1 $, and $ uv^i x y^i z \in L_3 $ for all $ i \geq 0 $.
Since $ |vxy| \leq p $, the substring $ vxy $ lies entirely within the first $ 0^p $, or spans $ 0^p 1 $, or lies in the second $ 0^p $. In every case, pumping (e.g., with $ i = 0 $) yields a string not of the form $ ww $, contradicting the assumption. Hence,
$$
L_3 \notin \text{CFL}.
$$
$$
\begin{array}{c|c}
\text{Language} & \text{Class} \\
\hline
L_1 & \text{CFL} \setminus \text{DCFL} \\
L_2 & \text{DCFL} \\
L_3 & \text{Not CFL} \\
\end{array}
$$
$$
\color{lime} \boxed{\text{Answer: B}}
$$
