If I say $X = \{x, x'\}\subset\mathbb{F}^3$ is a subspace, where $x$ and $x'$ are linearly independent (for some field $\mathbb{R}$ or $\mathbb{C}$), with
$$\mathbb{F}^n := \{(z_1,\dots,z_n)~:~z_1,\dots,z_n\in\mathbb{F}\},$$
and lets say I want to find a basis for $X^{\perp}$, that is, a basis for
$$X^{\perp}:=\{v\in \mathbb{F}^3~:~\langle x,v \rangle =0~\forall~x\in X\},$$
then with some sufficient work I should be able to find $X^{\perp}$, which has dimension $1$ since
$$\text{$X$ is a subspace of $\mathbb{F}^3$, so $\dim (\mathbb{F}^3)=\dim (X) + \dim (X^{\perp})$},$$
or $\dim(\mathbb{F}^3)-\dim(\{x, x'\})=3-2=1\dim(X^{\perp})$, that is to say, I can find such an element in $\mathbb{F}^3$, say $\zeta$, so that the system
$$\langle x,\zeta \rangle = \sum^3_{i=1}x_i\bar{z_i}=0$$ $$\langle x',\zeta \rangle = \sum^3_{i=1}x'_i\bar{z_i}=0$$
is satisfied. Here, $x_i$ means the $i^{th}$ element in $x=(x_1,x_2,x_3)\in X$, as is the case with $x'$. My question is, If I can find such a $\zeta$, then have I found the desired basis? If not, once I find such a $\zeta$, then how do I find a basis? This is me exploring the ideas discussed here.
Yes, a set containing a vector $\zeta=(z_1,z_2,z_3)$ is a basis for $X^{\perp}$. The matrix of your system will have rank 2 so the solution you get will be indeed onedimensional.