According to wikipedia a plane can be expressed as: $$(p-p_{0})\cdot n=0$$ where $$n - normal$$$$p_{0} - point$$however nothing is told about p. What is it? As I understand to define a plane I only need a vector and a normal.
Also how to solve it to get implicit formula of the plane?:$$Ax+By+Cz+D=0$$
On
Let $p_0=(a_0,b_0,c_0)$ and $p$ be any point on the plane $(x,y,z)$ Then $$\vec p-\vec p_0=(x-a)\hat i+(y-b) \hat j+(z-c)\hat k$$ And if the normal be $\vec n =n_x \hat i+n_y \hat j+n_z\hat k$ then you get an equation by $$(\vec p-\vec p_0)\cdot \vec n=0$$ i.e.
$$n_x(x-a)+n_y(y-a)+n_z(z-a)=0\\\implies A=n_x\ ;B=n_y\ ;C=n_z\ ;D=-(n_xa+n_yb+n_zc)$$
$p$ is the point you want to test and see if it belongs to the plane.
Let $n = (n_x, n_y, n_z)$ be a vector perpendicular to the plane and let $P_0$ be some point on the plane. Then the point $P$ belongs to the plane iff the vector $P - P_0$ is perpendicular to $n$. In terms of dot product, that is
$$(P - P_0)\cdot n = 0$$
If you write out explicitly the coordinates of the vector $P - P_0$ and expand the dot product, then you get something of the form $Ax + By + Cz = D$:
$$\begin{cases}P_0 = (x_0, y_0, z_0)\\P - P_0 = (x - x_0, y - y_0, z - z_0)\end{cases}\\ (P - P_0)\cdot n = (x - x_0, y - y_0, z - z_0)\cdot(n_x, n_y, n_z) = \\ n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0 $$
Collecting all the terms with the subscript $0$ gives the constant $D$. That is the reason why if you have a plane $Ax + By + Cz + D = 0$, then $(A, B, C)$ is a normal vector to the plane.