Find subspace $T$ of space $\mathbb R^3$ so that $\mathbb R^3=S \oplus T$ I have one problem. I am sure it is not complicated, but I only need help to see am I, at least, on the right path.
Problem: Let $S=Span\{(0,-2,3),(1,1,1),(2, -2, 8)\}\subseteq \mathbb R^3$. Find subspace $T$ of space $\mathbb R^3$ so that $\mathbb R^3=S \oplus T$.
Here is what I have done so far:


*

*Since $S$ is span of vectors $(0,-2,3),(1,1,1),(2, -2, 8)$, that means that $S$ has all vectors that are linear combination of those three vectors.

*We are searching for subspace $T$, but we need to keep in mind that $\mathbb R^3=S \oplus T$, which means that S$\cap T=\overrightarrow 0$. So, $T$ would have all those vectors that cannot be a result of linear combination of vectors from $S$.

*After that, I placed vectors from $S$ into a matrix:\begin{bmatrix}
        0 & 1 & 2 \\
       -2 & 1 & -2 \\
        3 & 1 & 8 \\
        \end{bmatrix} and I found its rank is $2$ which means that $\dim S=2$.We also know that $\dim\mathbb R^3=3$. Now, based on formula $\dim\mathbb R^3=\dim(S\oplus T)=\dim S+\dim T$, we get that dimension of $T$ should be $1$. That would mean that $T$ needs to have, of course, $3$ vectors, but the rank of matrix $[T]$ should be one. Am I right to assume those $3$ vectors would be something like 
$$T=\{(\alpha_0 a,\alpha_0 b,\alpha_0 c),(\alpha_1 a, \alpha_1 b, \alpha_1 c),(\alpha_2 a, \alpha_2 b, \alpha_2 c)\}$$, where $\alpha_0, \alpha_1$ and $\alpha_2$  are scalars? For example: $$[T]=\begin{bmatrix}
    2 & 7 & 1 \\
    4 & 14 & 2 \\
    8 & 28 & 4 \\
    \end{bmatrix}$$
Rank of that matrix would be one, making dimension of $T$ to be one. So, $T$ is span over one vector.


I searched here and found a similar problem but I guess I am not sure of how $T$ would really look like. Would $T$ be $span$ over vector $(1,0,0)$ because span over that vector cannot produce any in space $S$? Can $T$ be span over any vector that is making a base in $\mathbb R^3$? For example $T=span${$(1,1,0)$}?
Thank you.
Edit: changed last question.
 A: An idea: indeed, $\;\dim\mathcal S=\dim\text{Span}\,S=2\;$ ,so why won't you reduce your matrix (say, by rows to make it easier) to check what vector to take out (the one lin. dep. in the other two) and begin to check what vector to add in order to make the whole thing linearly independent?:
$$\begin{pmatrix}1&1&1\\2&-2&8\\0&-2&3\end{pmatrix}\stackrel{R_2-2R_1}\rightarrow\begin{pmatrix}1&1&1\\0&-4&6\\0&-2&3\end{pmatrix}\stackrel{R_3-\frac12R_2}\rightarrow\begin{pmatrix}1&1&1\\0&-4&6\\0&\;0&0\end{pmatrix}$$
and we see the third vector is lin. dep. in the first two, so you can add for example vector $\;(0,0,1)\;$ , and then
$$R^3=\mathcal S\oplus T\;,\;\;\text{with}\;\;T:=\text{Span}\,(0,0,1)$$
You can also choose $\;(0,1,0)\;$ instead, or infinite different vectors that'll work (how to know what? You can take a general vector $\;(a,b,c)\;$ and put it in the above matrix instead of the third row, and again reduce the matrix and check when the third row doesn't become all zeros)
Observe: $\;T\;$ does not need "to have three vectors", whatever that means: it has to be, by what we did above, a one-dimensional subspace generated, of course, by one single non-zero vector.
A: Forming a matrix with the vectors in $S$ as column vectors (as you have done), we get
\begin{equation*}
A = \begin{bmatrix}
0 & 1 & 2\\
-2 & 1 & -2\\
3 & 1 & 8
\end{bmatrix}.
\end{equation*}
Now, the span of $S$ is the same as the column space of $A$, and we can find a basis for this from the column echelon form of $A$, which is
\begin{equation*}
\begin{bmatrix}
1 & 0 & 0\\
1 & -2 & 0\\
1 & 3 & 0
\end{bmatrix}.
\end{equation*}
Now, there are two linearly independent vectors, which form a basis of the span of $S$. To get a complete basis of $\mathbb R^3$, you need one more vector, which is linearly independent from these two, and an obvious choice is (say) $v = \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$. Therefore, $T = \operatorname{span}(\{v\})$.
