Explanation:

The language {0n1n2n| 1 ≤ n ≤ 106} is a language of strings containing equal numbers of 0s, 1s, and 2s such that the number of 1s is between 1 and 106.

To determine whether this language is regular, context-free, or not context-free, we can use the pumping lemma for each of these classes of languages.

1. Regular:

Let L be a regular language. According to the pumping lemma for regular languages, there exists a constant p such that any string s in L of length at least p can be divided into three parts, s = xyz, such that:

1. |xy| ≤ p
2. |y| > 0
3. xyiz ∈ L for all i ≥ 0

We can apply this lemma to the language {0n1n2n| 1 ≤ n ≤ 106} by assuming that it is regular and deriving a contradiction.

Let p be the constant given by the pumping lemma for regular languages. Consider the string s = 0p1p2p ∈ L, which has length 3p. By the pumping lemma, s can be divided into three parts, s = xyz, such that |xy| ≤ p, |y| > 0, and xyiz ∈ L for all i ≥ 0.

Since |xy| ≤ p, the string y consists entirely of 0s. Therefore, if we pump y, we obtain a string that contains more 0s than 1s, which is not in L. This contradicts the pumping lemma, so the language {0n1n2n| 1 ≤ n ≤ 106} is not regular.

2. Context-free:

Let G = (V, Σ, R, S) be a context-free grammar for a language L. According to the pumping lemma for context-free languages, there exists a constant p such that any string s in L of length at least p can be divided into five parts, s = uvxyz, such that:

1. |vxy| ≤ p
2. |vy| > 0
3. uvixyiz ∈ L for all i ≥ 0

We can apply this lemma to the language {0n1n2n| 1 ≤ n ≤ 106} by assuming that it is context-free and deriving a contradiction.

Suppose there exists a context-free grammar G = (V, Σ, R, S) for the language {0n1n2n| 1 ≤ n ≤ 106}. We can construct a string s ∈ L that has length greater than p, where p is the constant given by the pumping lemma for context-free languages.

Let s = 0p1p2p ∈ L, which has length 3p. By the pumping lemma, s can be divided into five parts, s = uvxyz, such that |vxy| ≤ p, |vy| > 0, and uvixyiz ∈ L for all i ≥ 0.

Since |vxy| ≤ p, the substring vxy must be contained entirely within the first p symbols of s, which consists only of 0s. Therefore, we can write vxy = 0k for some integer k > 0.

If we pump vxy, we obtain a string that contains