Equation of a plane through three points ($(\mathbf{r}-\mathbf{a})\cdot [(\mathbf{b}-\mathbf{a})\times (\mathbf{c}-\mathbf{a})]=0$)

For three non-collinear points with position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, a nonzero normal vector isThe plane is the set of $\mathbf{r}$ satisfying $(\mathbf{r}-\mathbf{a})\cdot \mathbf{N}=0$. Expanding the cross product and using the scalar triple product givesNon-collinearity is essential: otherwise $\mathbf{N}=0$ and the equation fails to select a unique plane.