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?
Once you have $H$, to find $G$ you need to find a set of $k$ rows of length $n$ that are LI and orthogonal to the rows of $H$.
In a simple case like this, it could be done by trying, eg:
$$ G=
\begin{bmatrix}
1 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 \\
\end{bmatrix}$$
A more general and systematic way is to try to find an equivalent code (same codebook, different mapping) by doing elementary operations with the rows of $H$, to put it in systematic form: $H=( I | P´)$ and then $G=( P | I)$ fits the bill.
In this case, we cannot do that manipulating the rows alone. We can resort to an aditional trick: you are also allowed to permute some columns , to bring it to systematic form, but then at the end un-permute the rows in the resulting $G$.
So, let's permute columns 2 and 4 to get the modified parity matrix
$$\begin{bmatrix}
1 & 0 & 1 & 1 & 0 \\
0 & 1 & 0 & 0 & 1 \\
\end{bmatrix}=[I | P']
$$
and the modified generator matrix is
$$[P | I] =\begin{bmatrix}
1 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 1 \\
\end{bmatrix}
$$
and after permuting columns 2 and 4 we get
$$ G = \begin{bmatrix}
1 & 0 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 1
\end{bmatrix}
$$
You can (you should) check that the rows are indeed LI and orthogonal to the rows of $H$.
Best Answer
If you have a $k \times n$ generator matrix $G$ and message vector $m$ of length $k$, you can encode the message (i.e., find the code vector corresponding to that message vector) by computing $mG$.
To find all code vectors, simply repeat this computation for every possible value of $m$. As you observed, there are $2^k$ possible values of $m$ if you're working in binary.