Plane determined by $2$ vectors.

The plane determined by two noncollinear vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ is the collection of all vectors of the form $\alpha\mathbf{v}_1 + \beta\mathbf{v}_2$, with $\alpha$ and $\beta$ scalars.

If by "space" you happen to mean $\mathbb{R}^3$, $\mathbf{v}_1=(a,b,c)$ and $\mathbf{v}_2=(r,s,t)$, then $(a,b,c)\times(r,s,t)$ (the cross product) is perpendicular to both $(a,b,c)$ and $(r,s,t)$, hence perpendicular to the plane they determine, so it will be the normal to the plane. One you have the normal, presumably you know how to get the usual equation of the plane.

Take their cross product. This will be perpendicular to the plane they determine.