The question is very confusing. You say you have 8 codewords. Then you say you want to find a linear code. What, if any, is the connection between the 8 codewords you have and the linear code you want to find?
Then you say you want to use a $(7,4)$ Hamming code. But a $(7,4)$ Hamming code has 16 codewords, so how does this relate to the 8 codewords you say you have?
Then you say you need to find a generator matrix and a parity check matrix. Do you need these for the $(7,4)$ Hamming code? or is this for the 8 codewords you have?
The only part of the question I feel comfortable answering is, yes, you can find a generator matrix from a parity check matrix. Any good text or notes on coding theory should show you how to do that.
Then in the comments you ask whether there is an easier way than just using $(111000000)$, $(000111000)$, and $(000000111)$. I don't know if there is an easier way, since I can't figure out what you are trying to accomplish (see my first three paragraphs). But you can certainly use those vectors to form the generator matrix for an 8-word, 1-error-correcting linear code and, while there may be better ways, I can't imagine a simpler one.
It seems to me that questions about forming generator matrices and turning parity check matrices into generator matrices were answered in your earlier question, Coding Theory and Generating a matrix.
You don't need the Generator Matrix to decode to any codeword. What you can do is the following. The syndrome vector $s=r*H$ where $r$ is a received $1\mathrm{X}N$ vector. The syndrome can be identified also as follows: Let the received vector be $r=c_{t}+e$ where $c_{t}$ is the transmitted codeword and $e$ is the error vector that corrupted the codeword. We all know that the Parity check matrix is the null space for the codewords. Now then $s = e*H$ and you have the position of the codeword which had an error in it. Say if $s=e*H$ picked up the 3rd column of $H^{T}$, then the 3rd bit was erreneous in the received vector $r$.
Best Answer
Say $C$ is your code with generator matrix $G$. If you reduce $G$ to echelon form, you obtain $$\begin{bmatrix} 1&0&1&0&1&0&1\\0&1&1&0&0&1&1\\0&0&0&1&1&1&1\end{bmatrix}$$ which is unfortunately not in standard form.
BUT, we can put it in standard form by swapping the third and fourth column, so we get $$G^{\prime} = \begin{bmatrix} 1&0&0&1&1&0&1\\0&1&0&1&0&1&1\\0&0&1&0&1&1&1\end{bmatrix}$$
This $G^{\prime}$ is the generator matrix for a different (but equivalent) code $C^{\prime}$, where the 3rd and 4th positions of our codewords have been swapped. So $C^{\prime}$ has parity check matrix $$H^{\prime} = \begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 1 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 0 & 0 & 0 & 1 \end{bmatrix}$$ And we translate it back to a parity check for $C$ be swapping the third and fourth columns back again $$H = \begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 1 & 0 & 1 & 0 & 1 & 0\\ 1 & 1 & 0 & 1 & 0 & 0 & 1 \end{bmatrix}$$