Derive equation of plane through three points In my book it is given that

Equation of a Plane through three points
Vector form: If $A,B,C$ are three points having P.V.'s $\vec a,\vec b,\vec c$ respectively, then vector equation of the plane is $\begin{bmatrix}\vec  r&\vec a&\vec b\end{bmatrix}+\begin{bmatrix}\vec r&\vec b&\vec c\end{bmatrix}+\begin{bmatrix}\vec r&\vec c&\vec a\end{bmatrix}=\begin{bmatrix}\vec a&\vec b&\vec c\end{bmatrix}$
Cartesian form: The equation of the plane through three non-collinear points $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ is $\begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix}=0$

How have they written the equation of the plane? Can anyone derive it?

That vector equation is the expanded form of
$$
[(\vec b-\vec a)\times(\vec c -\vec a)]\cdot(\vec r-\vec a)=0.
$$
This is the equation of the plane through $ABC$, because 
$(\vec b-\vec a)\times(\vec c -\vec a)$ is a vector normal to the plane.
