v1 = (2 -1 2 1) v2 = (0 -1 1 -3) These vectors are column vectors.
Find a 2x4 matrix A, with linearly independent rows, such that Av1 = 0 and Av2 = 0.
Or, find a, b, c, d, e, f, g, h such that
$\displaystyle \begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}2 \\ -1 \\ 2 \\ 1\end{bmatrix}= \begin{bmatrix} 2a- b+ 2c+ d \\ 2e- f+ 2g+ h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$
and
$\displaystyle \begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}0 \\ -1 \\ 1 \\ -3\end{bmatrix}= \begin{bmatrix} -b+ c- 3d \\ -f+ g- 3h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$
That gives you four equations to solve for 8 values. Of course, there will be an infinite number of such matrices just as there are an infinite number of possible bases for the orthogonal space FernandoRevilla suggests.
You are right, but we only need two equations and four unknowns.
$\displaystyle <v_1,v_2>^{\perp}\equiv\begin{Bmatrix} 2x_1-x_2+2x_3+x_3=0\\{}\quad -x_2+x_3-3x_4=0 \end{matrix}$
has dimension $\displaystyle 2$, so two linearly independent solutions of the system are "your" $\displaystyle (a,b,c,d)$ and $\displaystyle (e,f,g,h)$ , that is, we needn't to denote twice the unknowns.
The subspace of $\displaystyle \mathbb{R}^4$ whose elements $\displaystyle x=(x_1,x_2,x_3,x_4)$ are orthogonal to $\displaystyle v_1$ and $\displaystyle v_2$. Equivalently:
$\displaystyle \begin{Bmatrix}(x_1,x_2,x_3,x_4)\cdot (2,-1,2,1)=0\\(x_1,x_2,x_3,x_4)\cdot (0,-1,1,-3)=0\end{matrix}\Leftrightarrow\begin{Bmatrix}2x_1-x_2+2x_3+x_4=0\\{}\quad\; \;-x_2+x_3-3x_4=0\end{matrix}$