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
$$ \left( \begin{array}{cc} 0 \\0\\1\\0\\0 \end{array} \right) \ \ , \ \ \left( \begin{array}{cc} 0 \\0\\0\\1\\0 \end{array} \right) \ \ , \ \ \left( \begin{array}{cc} 0 \\0\\0\\0\\1 \end{array} \right) \ \ . $$
Notice that:
$$\begin{cases}
a_1+a_2+a_3=0\\
a_4+a_5+a_6=0
\end{cases}$$
is a system of $2$ equations and $6$ variables. We can introduce $6-2 = 4$ real parameters, $x$, $y$, $z$ and $w$, such that:
$$\begin{cases}
a_1 = -x-y\\
a_2 = x\\
a_3 = y\\
a_4 = -z-w\\
a_5 = z\\
a_6= w
\end{cases}.$$
Therefore, all matrices $A$ which satisfy you requirements are in the form:
$$\begin{bmatrix}-x-y&x&y\\-z-w&z&w\end{bmatrix},$$
for given parameters $x, y, z$ and $w$.
Best Answer
Saying that there are "four independent variables" does not constitute a subspace. For example, the subset of all vectors in $\Bbb{R}^2$ such that $a_1+a_2+1=0$ is not a subspace despite there being an independent variable. Instead, you should use the subspace test.
Let $A, B \in V$ and $c \in \Bbb{R}$. We must show that: $$A+cB \in V$$ This can be otherwise stated as: $$(A+Bv)\pmatrix{1\\2\\3}=\pmatrix{0\\0}$$ Distribute on the left-hand side to get: $$(A+Bv)\pmatrix{1\\2\\3}=A\pmatrix{1\\2\\3}+cB\pmatrix{1\\2\\3}$$ Now, we are given $A, B \in V$, so obviously, those matrices times $\pmatrix{1\\2\\3}$ are $\pmatrix{0\\0}$: $$(A+Bv)\pmatrix{1\\2\\3}=\pmatrix{0\\0}+c\pmatrix{0\\0}=\pmatrix{0\\0}$$ This proves the equation we wanted, so $V$ is a subspace.
Now, to find the dimension, we set up what you did the first time: $$\text{Let }A=\pmatrix{a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6}$$ $$\pmatrix{a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6}\pmatrix{1\\2\\3}=\pmatrix{0\\0}$$ $$a_1+2a_2+3a_3=0 \\ a_4+2a_5+3a_6=0$$ If we express $A$ as a column vector with $a_i$s rather than a matrix, the above corresponds to the following matrix-vector product equation: $$\pmatrix{1 & 2 & 3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 & 3}\pmatrix{a_1\\a_2\\a_3\\a_4\\a_5\\a_6}=\pmatrix{0\\0}$$ Now, basically, we are finding the dimension of the null space of the matrix on the left. There are two pivot columns and six columns in all, giving us a dimension of $4$.