I have the following set of solutions which is a subspace of $\Bbb{R}^4$
$$ \textsf{W} = \operatorname{span}\big(\{(1, 1, 2, 3), (1, 4, 2, 1)\}\big) $$
I am trying to find a system of homogeneous linear equations such that its solutions set is the given subspace. I think that the following matrix will produce the right solution, since the vector $\vec{x}=(x_1,x_2,x_3,x_4)$ has to be a linear combination of the given two vectors that are inside $\textsf W$.
$$ A= \begin{pmatrix} 1 & 1 & x_1 \\ 1 & 4 & x_2 \\ 2 & 2 & x_3 \\ 3 & 1 & x_4 \\ \end{pmatrix} $$
I know that the solution would basically depend on those rows that end up having a form similar to $0 = ax_2 + bx_3 $ for example, but I am not sure why that is. If someone could clarify, I would be very grateful.
In the equation $a_1x_1+a_2x_2+a_3x_3+a_4x_4=0$ the fixed coefficients $a_i$ and the unknowns $x_i$ play symmetric roles. Given a set of vectors that span the space defined by this equation, then, you can instead treat the $a_i$ as the unknowns and the $x_i$ as the fixed coefficients and generate a system of equations for them. There will, of course, be an infinite number of solutions.
Another way to look at this is to write the above equation as $\mathbf a\cdot\mathbf x=0$. The solution set is therefore the set of vectors orthogonal to $\mathbf a$, and for a system of equations $\mathbf a_k\cdot\mathbf x_k=0$, the vectors that satisfy it must be orthogonal to all of the $\mathbf a_k$, i.e., they are the orthogonal complement of $\operatorname{Span}\{\mathbf a_k\}$. As before, $\mathbf a$ and $\mathbf x$ play symmetric roles, so to find the coefficients for a system of equations with solution set equal to the span of the $\mathbf x_k$, you need to find a spanning set for its orthogonal complement.
Putting this into practice for your specific problem, following either method leads to computing the null space of $$\begin{bmatrix}1&1&2&3\\1&4&2&1\end{bmatrix}.$$