Write your four vectors as column vectors of a $ \ 5 \times 4 \ $ matrix and row reduce it:
$$ \left( \begin{array}{cc} 6 & 1 & 1 &7 \\4&0&4&1\\1&2&-9&0\\-1&3&-16&-1\\2&-4&22&3 \end{array} \right) \ \ \rightarrow \ \ \left( \begin{array}{cc} 0 & 0 & 0 &0 \\0&0&0&0\\0&0&0&1\\1&-3&16&1\\0&1&-5&\frac{1}{2} \end{array} \right) $$
[I used the fourth row here to work against the other rows; that doesn't matter particularly.]
What does the "zeroing-out" of two rows tell us? How can we use what non-zero rows remain to construct a basis for span(S) ? (Notice that these are five-dimensional vectors, so we are already starting out "short a coordinate variable", making it "free".)
EDIT -- Since the discussion has advanced further, we can say something about the basis of span(S). Taking the hint from Omnomnomnom or the above, the subspace spanned by your set of four vectors only has dimension 3. So we need to set up three linearly independent vectors, using the columns of the row-reduced matrix.
We could "reduce" those last two rows a bit more to obtain
$$ \rightarrow \ \ \left( \begin{array}{cc} 0 & 0 & 0 &0 \\0&0&0&0\\0&0&0&1\\1&0&1&0\\0&1&-5&0 \end{array} \right) \ \ . $$
With the matrix fully "reduced", we need to pick out three (five-dimensional) column vectors which are linearly independent. The third column is a linear combination of the first two, so we can toss that one out. A suitable basis for span(S) is then
$$ \left( \begin{array}{cc} 0 \\0\\1\\0\\0 \end{array} \right) \ \ , \ \ \left( \begin{array}{cc} 0 \\0\\0\\1\\0 \end{array} \right) \ \ , \ \ \left( \begin{array}{cc} 0 \\0\\0\\0\\1 \end{array} \right) \ \ . $$
Since you know $x$ and $y$ must satisfy the relation $x + y = 0$, then you can rewrite your restriction as: $$x = -y$$
And your vector space will be defined as: $(-y, y, z)$. Now, there are several ways to justify that this vector space is bidimensional. I believe that the most simple one is to notice that it is defined by only two independent variables, so you need only two coordinates to uniquely define a vector in this vector space, thus making it bidimensional.
You can also show this by finding the canonical base to this vector space, and showing that it has only two vectors - $(-1, 1, 0)$ and $(0, 0, 1)$.
The geometrical meaning of a subspace of a three dimensional space being a two dimensional space is that all the vectors from that subspace are contained on a plane in the three dimensional space - besides the meaning of needing only 2 coordinates do be uniquely defined even on a three dimensional space, because the third coordinate is defined as a function of those two.
Best Answer
Hint:
Write the coordinates of the vectors as row vectors of a matrix, and perform row reduction of this matrix. The rank of the matrix will be the dimension of the subspace, and the non-zero lines in the final matrix will correspond to the vectors that span the subspace.