If I have the code $C=\{0000, 1110, 0212, 2011, 0121,1022, 2220,1201,2102\}\subset F^{4}_3$ is correct to say The code is linear over the finite subespace $F_3$ with $n=4$
$|C|=9$ and $k=dimC=\log_3{9}=2$ so $C$ is a code $[4,2]-linear$
The Generate matrix have $k=2$ lines and $n=4$ columns
$G=\left(\begin{array} 11110 \\ 0212\end{array}\right)$
$d\left(C\right)=3$
Is that corret? How can I find de Parity matrix and the dual Code?
The code is self dual. Going just by the definition, a vector $v$ will be in the dual if $v \cdot w^T = 0$ for every $w$ in our original code. Since our code consists of all linear combinations of $\begin{bmatrix}1&1&1&0 \end{bmatrix}$ and $\begin{bmatrix}0&2&1&2 \end{bmatrix}$ we just need $$\begin{bmatrix}v_1&v_2&v_3&v_4 \end{bmatrix} \begin{bmatrix}1\\1\\1\\0 \end{bmatrix} = \begin{bmatrix}0 \end{bmatrix}$$ and $$\begin{bmatrix}v_1&v_2&v_3&v_4 \end{bmatrix} \begin{bmatrix}0\\2\\1\\2 \end{bmatrix} = \begin{bmatrix}0 \end{bmatrix}.$$
This tells us we want $$v_1+v_2+v_3=0$$ and $$2v_2+v_3+2v_4=0$$ which gives us a linear system of equations. We can put this into a matrix and solve $$\begin{bmatrix}1&1&1&0&0\\0&2&1&2&0\end{bmatrix}.$$ We can row reduce, doing everything in $\mathbb{Z}_3$ keeping in mind that $-1=2$ here to get $$\begin{bmatrix}1&0&2&2&0\\0&1&2&1&0\end{bmatrix}.$$ This means the solutions satisfy $x_1 = x_3+x_4$ and $x_2=x_3-x_4$. So we can write $$\begin{bmatrix}x_3+x_4\\x_3-x_4\\x_3\\x_4\end{bmatrix}= \begin{bmatrix}x_3\\x_3\\x_3\\0\end{bmatrix}+\begin{bmatrix}x_4\\-x_4\\0\\x_4\end{bmatrix}=x_3 \begin{bmatrix}1\\1\\1\\0\end{bmatrix}+x_4 \begin{bmatrix}1\\2\\0\\1\end{bmatrix}.$$
Our dual code is therefore the set of all linear combinations of $\begin{bmatrix}1&1&1&0 \end{bmatrix}$ and $\begin{bmatrix}1&2&0&1 \end{bmatrix}$. Subtracting the former from the latter gives us the set of linear combinations of $\begin{bmatrix}1&1&1&0 \end{bmatrix}$ and $\begin{bmatrix}0&1&2&1 \end{bmatrix}$. Note that $\begin{bmatrix}0&1&2&1 \end{bmatrix}$ is just twice $\begin{bmatrix}0&2&1&2 \end{bmatrix}$ which shows that the dual is just the set of the exact same linear combinations as our original code. As your other question has been handled in the comments, we can stop here.