Deriving the equation of a plane $Ax+By+Cz=D$. Defining a plane as the span of two linearly independent vectors, I've been trying to derive the equation $$Ax+By+Cz=D$$ without much success. The equation seems to indictate that a vector
$$\vec{v}=\begin{bmatrix} x \\ y \\ z\end{bmatrix}$$
is in the plane if and only if $Ax+By+Cz=D$.

I was wondering if anyone could at least point me in the right direction as to how to prove the two definitions are equivalent.
 A: The span of vectors contains the origin, but in general
$$A\,0+B\,0+C\,0\ne D.$$

A plane can be defined as the affine set
$$\vec p=\lambda\vec a+\mu\vec b+\vec c.$$
We can eliminate $\lambda,\mu$ by forming the dot product with $\vec a\times\vec b$,
$$\vec a\times\vec b\cdot\vec p=\vec a\times\vec b\cdot\vec c,$$
which is of the form
$$Ax+By+Cz=D.$$
A: A plane containing the origin is the span of two linearly independent vectors $\vec{a}$ and $\vec{b}$.  The plane normal is $\vec{n} = \vec{a}\times \vec{b}$.  If $\vec{x}$ is a point of the plane, $\vec{x}=u\vec{a} + v\vec{b}$ and $\vec{n}\cdot \vec{x} = 0$ since $\vec{n}\cdot \vec{a}=0$ and $\vec{n}\cdot \vec{b}=0$.  To transform the equation $\vec{n}\cdot\vec{x}=0$ to the more general equation, simply apply a translation.  More specifically, $\vec{n}\cdot (\vec{x}-\vec{P}) = 0$ is the equation of a plane containing the point $P$.
A: Consider the linear map $f: \mathbb{R}^3 \to \mathbb{R}$ defined by $f(x,y,z) = Ax + By + Cz$. If $A, B, C$ are all zero, then this is just the zero map, so the kernel of $f$ is all of $\mathbb{R}^3$. If at least one of $A, B, C$ is nonzero, then the image spans $\mathbb{R}$ and has dimension 1, so by the rank-nullity theorem the kernel has dimension 2, i.e. it is a plane.
The kernel of $f$ is the set of all $(x,y,z)$ such that $Ax + By + Cz = 0$. Therefore the equation $Ax + By + Cz = D$ is a translation of the kernel of $f$. Specifically, the set of all vectors satisfying the equation $Ax + By + Cz = D$ is the plane $v + \ker f$, where $v$ is any vector in the plane.
We can see geometrically how this characterizes the set of all planes in $\mathbb{R}^3$: Given a plane in $\mathbb{R}^3$, it is just the translation of some plane containing the origin, i.e. it is just a translation of some subspace of dimension 2. So any plane can be written as $v + U$ for some vector $v$ and some subspace $U$ of dimension 2. $U$ can be expressed as the kernel of some linear map from $\mathbb{R}^3$ to $\mathbb{R}$.
A: Hint
If a point $M=(x,y,z)^T$ belongs to a plane passing by $O=(O_x,O_y,O_z)^T$ and spanned by the two vectors $v_1,v_2$, then the determinant $\det(OM,v_1,v_2)$ vanishes.
A: Another way : 
Plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$ :
$$\overrightarrow{a} = \left[\begin{matrix} p \\ q \\ r  \end{matrix}\right] \ and \ \overrightarrow{b} = \left[\begin{matrix} l \\ m \\ n  \end{matrix}\right]$$
Now, define a vector $\overrightarrow{v}$ which goes from $(x_1,y_1,z_1)$, to a point on the plane and $(x,y,z)$, any arbitrary point on the plane.
$$\overrightarrow{v} = \left[\begin{matrix} x-x_1 \\ y-y_1 \\ z-z_1 \end{matrix}\right]$$ 
So since $\overrightarrow{v}$, $\overrightarrow{a}$ and $\overrightarrow{b}$ are coplanar. So,
$$ \left[\begin{matrix} \overrightarrow{a} & \overrightarrow{b} & \overrightarrow{v} \end{matrix}\right] = 0$$
$$\begin{vmatrix} p & q & r \\ l & m & n \\ x-x_1 & y-y_1 & z-z_1 \end{vmatrix} = 0$$
Replace the constant coefficients with $A,B,C \&D$
$$Ax+By+Cz+D=0$$
