I assume that you want to find two linearly independent equations
$$ a_{11}x + a_{12}y + a_{13}z + a_{14}w = 0, \\
a_{21}x + a_{22}y + a_{23}z + a_{24}w = 0 $$
whose solution space is the vector space $W$. To do that, write a general linear equation
$$ ax + by + cz + dw = 0 $$
and plug in for $(x,y,z,w)$ a basis of $W$ yielding the equations
$$ a - 2b + 3d = 0, \\
a - b - c + 4w = 0. $$
Solving the equations for $(a,b,c,d)$ will give you a two-dimensional subspace of solutions and choosing any two linearly independent vectors $(a_{11}, a_{12}, a_{13}, a_{14})$ and $(a_{21}, a_{22}, a_{23}, a_{24})$ from the solution space will give you two linearly independent equations whose solution space is the vector space $W$.
There is an important high level concept called duality that can explain why this works. Solving a system of linear equations using Gauss eliminations allows you to pass from a subspace $W \subseteq \mathbb{F}^n$ described as a solution space of linear equations
$$ W = \{ v \in \mathbb{F}^n \, | \, \varphi_1(v) = \ldots = \varphi_k(v) = 0 \} $$
(here $\varphi \colon \mathbb{F}^n \rightarrow \mathbb{F}$ are linear functionals, not necessarily independent) to a description of $W$ of the form
$$ W = \mathrm{span} \{ v_1, \ldots, v_l \} $$
where $v_1, \ldots, v_l$ are vectors in $\mathbb{F}^n$ that form a basis for $W$. By treating the vectors $v_i$ as functionals on $(\mathbb{F}^n)^{*}$, one can see that the same procedure (under some identifications) allows you to pass from a subspace $W$ described as
$$ W = \mathrm{span} \{ u_1, \ldots, u_k \} $$
where the $u_i$ are (not necessarily linearly independent) vectors in $\mathbb{F}^n$ to a description of $W$ of the form
$$ W = \{ v \in \mathbb{F}^n \, | \, \varphi_1(v) = \ldots = \varphi_l(v) = 0 \} $$
where the $\varphi_i$ are linearly independent functionals (interpreted as equations) whose common solution space is precisely $W$.