 Solution

The set is the kernel of the linear map $L(\mathbf{r})=\mathbf{v}\cdot \mathbf{r}$. It contains zero, and linearity gives $L(\alpha \mathbf{r}+\beta \mathbf{s})=0$ whenever $L(\mathbf{r})=L(\mathbf{s})=0$, proving closure under linear combinations. If $\mathbf{v}\ne 0$, it is a plane through the origin; if $\mathbf{v}=0$, it is all of ${\mathbb{R}}^3$. An arbitrary plane need not be a vector subspace: for example $z=1$ omits the zero vector. An affine subspace and a vector subspace are not interchangeable.