Just so we're on the same page, here's the definition of control matrix I'm familiar with.
Definition: Let $C$ be a $[n,k]$ linear code over a field $F$ with generator matrix $G$, so that $C = \{ mG \mid m \in F^k \}$. If $H$ is a matrix such that
$$C = \{ x \in F^n \mid Hx^T = 0 \},$$
then $H$ is called a control matrix of $C$.
First, we need a small lemma. You may have already seen this in your class.
Lemma: Let $C$ be a $[n,k]$ linear code over a field $F$ with generator matrix $G$ and let $S$ be an invertible matrix. Let $D$ be the code with generator matrix $GS$. If $H$ is a control matrix of $D$, then $HS^T$ is a control matrix of $C$.
Proof: Since $D$ has generator matrix $GS$ and control matrix $H$, we have
$$ D = \{ mGS \mid m \in F^k \} = \{ x \in F^n \mid Hx^T = 0 \}. $$
Note that $C = \{ mG \mid m \in F^k \}$. Define $X = \{ x \in F^n \mid HS^T x^T = 0 \}$. We will show that $C = X$.
($C \subseteq X$) Consider any $c \in C$. Then $cS \in D$, so $HS^T c^T = H(cS)^T = 0$, and so $c \in X$.
($C \supseteq X$) Let $x \in F^n$ and $HS^T x^T = H(xS)^T = 0$. Then $xS \in D$, so there is some $m \in F^k$ such that $mGS = xS$. Since $S$ is invertible, we have $mG = x$, so $x \in C$.
Since $C = X = \{ x \in F^n \mid HS^T x^T = 0 \}$, we have that $HS^T$ is a control matrix for $C$.
Back to your problem. Let
$$
G = \begin{bmatrix}
0&0&0&0&1&1&1&1\\
0&0&1&1&0&0&1&1\\
0&1&0&1&0&1&0&1\\
1&1&0&0&1&1&0&0
\end{bmatrix}.
$$
Put $G$ in reduced row echelon form to get a matrix
$$
G' = \begin{bmatrix}
1&0&0&1&0&1&1&0\\
0&1&0&1&0&1&0&1\\
0&0&1&1&0&0&1&1\\
0&0&0&0&1&1&1&1
\end{bmatrix}.
$$ Note that the row space of $G$ and $G'$ are the same, so they're both generator matrices of $C$. The problem here is $G'$ isn't in the form $[I|P^T]$. Let $S$ be the $8 \times 8$ permutation matrix that swaps columns 4 and 5:
$$
S = \begin{bmatrix}
1&0&0&0&0&0&0&0\\
0&1&0&0&0&0&0&0\\
0&0&1&0&0&0&0&0\\
0&0&0&0&1&0&0&0\\
0&0&0&1&0&0&0&0\\
0&0&0&0&0&1&0&0\\
0&0&0&0&0&0&1&0\\
0&0&0&0&0&0&0&1\\
\end{bmatrix}.
$$
Then $G'S = [I|P]$ for some matrix $P$, so the code with generator matrix $G'S$ has control matrix $[-P^T|I]$. By the lemma, $[-P^T|I]S^T$ is then a control matrix for $C$.
TL;DR: Let $C$ have generator matrix $G$. Put $G$ in RREF. Swap columns until it's in the form $[I|P]$. Let $H = [-P^T|I]$. Swap the same columns in $H$ in the reverse order. The resulting matrix is a control matrix of $C$.
A "standard" generator matrix of a $[n,k]$ code usually means that we seek a $k\times n$ matrix $G$ of the form $$G = \left[I_{k\times k}\ P_{k\times (n-k)}\right]$$ where $I_{k\times k}$ denotes a $k\times k$ identity matrix. For a $[7,6,2]$ code, $P_{6\times 1}$ is just a column of $6$ bits, and I leave it to you to figure what the six bits must be in order for the code to have minimum distance $2$. Do it; write out an arbitrary 6-bit column on the right of a $6\times 6$ identity matrix. Then, step back and admire the $6\times 7$ matrix that you
have written down, and ponder the fact that each row of $G$ is a codeword.
Best Answer
So let's say your generator matrix is $G$, which you described as having three codewords (we'll say as rows) $c_1,c_2,c_3$ from top to bottom.
This thing is called the generator matrix because it produces every codeword as a result of multiplication by some vector in $\Bbb F_2^3$ on the left, like this: $xG=c$.
So, what are the possible inputs? From the context, I gather you are working over $\Bbb F_2$. So, the possible inputs are: $[0,0,0], [1,0,0], [0,1,0], [0,0,1],[1,1,0],[1,0,1],[0,1,1],[1,1,1]$.
The first seven correspond to words you already listed: $0, c_1,c_2,c_3,c_1+c_2,c_1+c_3,c_2+c_3$ in that order.
The codeword you were missing corresponds to the last one: $c_1+c_2+c_3$!
In general, if you have a code over $\Bbb F_2$ and a $k\times n$ generator matrix (that is, $k\leq n$, $n$ is the length of the code and $k$ is the dimension.) then all of the codewords will be given by multiplying by the vectors from $\Bbb F_2^k$. Since there are $2^k$ of these vectors, there will be $2^k$ codewords.
If instead you are over a larger finite field like $\Bbb F_q$, then the number of codewords will be $q^k$. Can you see why?