I have a plane at a point $\boldsymbol{r}_{0}\in\mathbb{R}^3$ with normal vector $\boldsymbol{n}\in\mathbb{R}^3$. $$ \boldsymbol{n}^\top (\boldsymbol{r} - \boldsymbol{r}_0) = 0 $$ My aim is to find a basis for this plane. I know that if this plane was through $\boldsymbol{r}_0 = \boldsymbol{0}$ then one way in which I could find a basis for the plane would be via SVD decomposition. The last $2$ columns of $U$ would form a basis for $N((\boldsymbol{n}^\top)^\top) = N(\boldsymbol{n})$, the null space of the "matrix" $\boldsymbol{n}$.
$$ \begin{pmatrix} n_1 & n_2 & n_3 \end{pmatrix} \begin{pmatrix} r_1 \\ r_2 \\ r_3 \end{pmatrix} = 0 $$
My question is:
How can I find a basis for the plane at $\boldsymbol{r_0}$ with normal vector $\boldsymbol{n}$ using SVD/QR decompositions? I DO NOT want a different method, I just want to adjust this method to find the correct basis.
Consider the plane at the origin, with basis $\{u_1, u_2\}$ and at some other point $r_0$. These are parallel. So $\{u_1, u_2\}$ serves as a basis for the second one as well.
Well...to be more honest, the second one isn't a linear subspace of $3$-space (it doesn't contain the zero-vector), so it doesn't even have a (linear) basis. But it has an affine basis, or "coordinate system", with origin $r_0$ and the two vectors from the origin-plane as vectors. Or it has an affine "frame", consisting of $r_0, r_0 + u_1, r_0 + u_2$.
Or if you want to consider it as lying in the tangent-space to $\Bbb R^3$ at $r_0$, it is a subspace of that, with basis $(r_0; u_1), (r_0; u_2)$, where I'm using the standard identification of $T\Bbb R^3$ with $\Bbb R^3 \times \Bbb R^3$ here.