The standard way is to form a matrix with these vectors as columns, which has column space equal to $\mathbb{R}^{5}$: $$\begin{bmatrix} 3&0&0&1&0&0&0&0 \\\ 1&0&0&0&1&0&0&0 \\\ 0&7&0&0&0&1&0&0 \\\ 0&1&0&0&0&0&1&0 \\\ 0&0&1&0&0&0&0&1 \end{bmatrix}$$ Now you want to follow the method to get a basis for the column space: Figure out which are the pivot columns, and use those as a basis for $\mathbb{R}^{5}$ (note that with this setup, since your original three vectors were linearly independent, they will be pivot columns and so used as the first three vectors of the basis).