Consider the code $C=\{0000, 0010, 0020, abcd, 2110, 2100, 1220, 2120, 1210\}\subset F^{4}_3$
To determine $abcd\in F^{4}_3$ for $C$ to be linear is to say that $abcd$ is a linear combination of a base of $C$
considering the vectors $0010 , 1220$ a base of $C$ and two constant $\alpha,\beta\in F_3=\{0,1,2\}$ so we have $abcd=\alpha\times 0010+\beta\times 1220\Rightarrow \begin{cases} a=\beta\\ b=2\beta\\ c=\alpha+2\beta\\d=0 \end{cases}$
If $\alpha=\beta=0$ we have a word $0000$
If $\alpha=0,\beta=1$ we have a word $1220$
If $\alpha=1, \beta=0$ we have a word $0010$
If $\alpha=0,\beta=2$ we have a word 2110$
If $\alpha=2,\beta=0$ we have a word $0020$
If $\alpha=1,\beta=2$ we have a word $2120$
If $\alpha=2,\beta=1$ we have a word $1210$
If $\alpha=1,\beta=1$ we have a word $1200$ If $\alpha=2,\beta=2$ we have a word $2100$
as $1200$ is the only word that is not in the code, then $abcd = 1200$
AM I RIGHT?
To make a code with nine words you need exactly two vectors. Pick any two linearly independent vectors and you immediately generate all nine code words. For example $0010$ and $2100$. I picked these due to the number of zeroes. The nonzero elements are in different spots, so $$a \times 0010 + b \times 2100 = (2b )ba0.$$ This gives us all words with 0 in the last spot, where the first is twice the second. Our list becomes $\{0000,0010,0020,2100,2110,2120,1200,1210,1220 \}.$ Just see which one is missing from the original list and you'll have the answer.