 Plane through a point and a line given by a cross product

Suppose $\mathbf{a}\ne 0$ and $\mathbf{a}\cdot \mathbf{b}=0$. The line $\mathbf{r}\times \mathbf{a}=\mathbf{b}$ has parametrization $\mathbf{r}={\mathbf{r}}_0+t\mathbf{a}$, where ${\mathbf{r}}_0=(\mathbf{a}\times \mathbf{b})/|\mathbf{a}{|}^2$. For a point $\mathbf{y}$ outside the line, a normal vector of the unique plane containing both is $\mathbf{N}=\mathbf{a}\times (\mathbf{y}-{\mathbf{r}}_0)=\mathbf{b}+\mathbf{a}\times \mathbf{y}$. Thus its equation is the one in the title. The vector triple product verifies ${\mathbf{r}}_0\times \mathbf{a}=\mathbf{b}$, and cyclic invariance of the scalar triple product verifies that every point of the line satisfies the equation. The excluded case $\mathbf{N}=0$ is exactly $\mathbf{y}$ lying on the line, when the plane is not unique.