Consider the languages $L1$, $L2$ and $L3$ as given below. $L1 = \{0^p 1^q \mid p, q \in \mathbb{N}\}$, $L2 = \{0^p 1^q \mid p, q \in \mathbb{N} \text{ and } p = q\}$ and $L3 = \{0^p 1^q 0^r \mid p, q, r \in \mathbb{N} \text{ and } p = q = r\}$. Which of the following statements is NOT TRUE?

Let us analyze the three languages:

1. $L1 = \{0^p 1^q \mid p, q \in \mathbb{N}\}$ represents the regular language given by the regular expression $0^*1^*$. Since every regular language is also context-free, $L1$ is regular and can be recognized by a finite automaton as well as a pushdown automaton (PDA).

2. $L2 = \{0^p 1^q \mid p, q \in \mathbb{N} \text{ and } p = q\} = \{0^n 1^n \mid n \in \mathbb{N}\}$ requires matching the number of $0$s with the number of $1$s. It is a deterministic context-free language (DCFL) and can be recognized by a PDA.

3. $L3 = \{0^p 1^q 0^r \mid p, q, r \in \mathbb{N} \text{ and } p = q = r\} = \{0^n 1^n 0^n \mid n \in \mathbb{N}\}$ requires comparing three independent counts simultaneously, which cannot be done with a single stack. By the pumping lemma for context-free languages, $L3$ is not context-free (it is context-sensitive / recursively enumerable).

4. Turing machines can recognize all recursively enumerable languages, which includes $L1$, $L2$, and $L3$.

Therefore, statement (C) ("All the three languages are context free") is NOT TRUE.