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Here the pumping lemma is used to find the language is context free or not.

$\begin{array}{c}\left|\mathrm{z}\right|鈮�\mathrm{n}\\ \mathrm{z}=\mathrm{uvwxy}\\ \left|\mathrm{v}\right|鈮�1\end{array}$

Here at least 1 is not null.

$\begin{array}{c}\mathrm{L}=\left\{{\mathrm{a}}^{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}{\mathrm{c}}^{\mathrm{n}}\right|\mathrm{n}鈮�0\}\\ \mathrm{z}=\mathrm{uvwxy},\\ \mathrm{z}={\mathrm{a}}^{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}{\mathrm{c}}^{\mathrm{n}}\{\mathrm{n}=1,\mathrm{z}=\mathrm{abc}\}\\ \left|\mathrm{z}\right|=3\mathrm{n}>\mathrm{n}\end{array}$

If we equate both z then we get,

$\begin{array}{c}\mathrm{uvwxy}={\mathrm{a}}^{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}{\mathrm{c}}^{\mathrm{n}}\\ \left|\mathrm{vx}\right|鈮�1,\{\mathrm{n}鈮�|\mathrm{vx}|鈮�1\}\end{array}$

Then after that,

Let v or x is a part of -

$\begin{array}{l}\left({\mathrm{a}}^{\mathrm{i}}{\mathrm{b}}^{\mathrm{j}}\right)\mathrm{or}\left({\mathrm{b}}^{\mathrm{i}}{\mathrm{c}}^{\mathrm{j}}\right)\\ \mathrm{v}=\left({\mathrm{a}}^{\mathrm{i}}{\mathrm{b}}^{\mathrm{j}}\right)\\ {\mathrm{v}}^2=\left({\mathrm{a}}^{\mathrm{i}}{\mathrm{b}}^{\mathrm{j}}\right)\left({\mathrm{b}}^{\mathrm{i}}{\mathrm{c}}^{\mathrm{j}}\right)\\ {\mathrm{uv}}^2{\mathrm{wx}}^2\mathrm{y}={\mathrm{a}}^{\mathrm{m}}{\mathrm{b}}^{\mathrm{m}}{\mathrm{c}}^{{}^{\mathrm{m}}}\end{array}$

From this equation, it is proof that

${\mathrm{uv}}^2{\mathrm{wx}}^2\mathrm{y}鈭�\mathrm{L}$

Therefore, L is not context free language.

Suppose$\mathrm{C}$is a context free language and Let $\mathrm{p}$be the pumping length and$\mathrm{s}=1^{\mathrm{p}}{3}^{\mathrm{p}}{2}^{\mathrm{p}}{4}^{\mathrm{p}}鈭�\mathrm{C}$.

Since, $\left|\mathrm{s}\right|鈮�\mathrm{p},$then write,

$\mathrm{s}=\mathrm{uvxyz}$

and satisfying the condition of the pumping lemma.

Since, $\left|\mathrm{vxy}\right|鈮�\mathrm{p},$

$\begin{array}{c}\mathrm{L}=\left\{{\mathrm{a}}^{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}{\mathrm{c}}^{\mathrm{n}}\right|\mathrm{n}鈮�0\}\\ \mathrm{z}=\mathrm{uvwxy},\\ \mathrm{z}={\mathrm{a}}^{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}{\mathrm{c}}^{\mathrm{n}}\{\mathrm{n}=1,\mathrm{z}=\mathrm{abc}\}\\ \left|\mathrm{z}\right|=3\mathrm{n}>\mathrm{n}\end{array}$

If we equate both z then we get,

$\begin{array}{l}\mathrm{uvwxy}={\mathrm{a}}^{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}{\mathrm{c}}^{\mathrm{n}}\\ \left|\mathrm{vx}\right|鈮�1,\{\mathrm{n}鈮�|\mathrm{vx}|鈮�1\}\end{array}$

Then after that,

there are three possible cases

1) If $\mathrm{vxy}$is substring of $1^{\mathrm{p}}{3}^{\mathrm{p}}$, then in ${\mathrm{uv}}^2{\mathrm{xy}}^2\mathrm{z}$, either the number of 1s is greater than the number of two鈥檚, or the number of three鈥檚 is greater than number of four, or both. Thus, uv2xy2 z $鈭�C$, which is a contradiction.

2) If$\mathrm{vxy}$ is substring of $2{}^{\mathrm{p}}4{}^{\mathrm{p}}$, then similar to ${\mathrm{uv}}^2{\mathrm{xy}}^2\mathrm{z}$$鈭�\mathrm{C}$, which is a contradiction.

3) If $\mathrm{vxy}$is a substring of $3{}^{\mathrm{p}}2{}^{\mathrm{p}}$ them in${\mathrm{uv}}^2{\mathrm{xy}}^2\mathrm{z}$ , either the number of twos is greater than the number of one鈥檚 or the number of three鈥檚 is greater than number of fours, or both. Thus,${\mathrm{uv}}^2{\mathrm{xy}}^2\mathrm{z}$$鈭�\mathrm{C}$ , which is a contradiction.