Why does a plane with normal constant vector pass through the origin? Given a vector $r=\langle r_1,r_2,r_3\rangle$ it is known that $n=\langle n_1,n_2, n_3\rangle$, a vector perpendicular to it, is constant. It transpires that the plane which contains the vector $r$ passes through the origin. 
The proof given for this is that $r\cdot n=0 \Rightarrow r_1n_1+r_2n_2+r_3n_3=0$ and $x=y=z=0$ satisfies the equation. 
Why does this prove that the plane passes through the origin?
Is $r\cdot n$ the plane equation? 
Doesn't any plane that go through the origin by that logic?
This answer was given in the comments to the answer of this question: Prove that a particle is traveling on a plane from its velocity and acceleration in space
This is the visualization:

Here the plane is drawn to pass through origin but I could've drawn it to pass through any other point for example.
$\mathbf{EDIT:}$ at the end the explanation is quite simple: any vector is defined to start at the origin. Because the position vector travels on a plane and its normal vector is constant then, the motion has to start at the origin. If someone wants to post an answer I will gladly accept it.
 A: The comments give one way of understanding, but I figured I could maybe explain the original proof better.
Suppose our plane has normal vector $n$ (which is normal to the plane at every point on it) and an example point on the plane: $p$. These two vectors, one a direction and the other a point, determine any plane.
Then, our plane equation is $$ n\cdot(x-p)=0 $$
that is, every $x$ that satisfies this equation is a member of the plane. This is equivalent to $n\cdot x = n\cdot p$.
Notice that $x=p$ is a member of the plane. But, $x=p+n$ is not a member, nor is $x=\vec{0}$, assuming $p\ne \vec{0}$.
Another way to write this is as a point set $\Pi$:
$$
\Pi = \{ \;x\in\mathbb{R}^3\;|\;n\cdot x= n\cdot p\;\}
$$
i.e. these are the set of points making up the plane. Again, notice that if $p=\vec{0}$, then the origin is a point in $\Pi$. But if $p$ does not vanish, then the origin is not in the plane. 
The reason is that $p$ is a translation or shifting parameter. That is, a plane has an orientation parameter, $n$, which "rotates" it, and a position parameter $p$, which slides the plane around. When $p=\vec{0}$, we have slid the plane so that it intersects that origin. The plane equation in this case is $n\cdot x = 0$.

Here's a different approach. Every plane is determined by giving 3 unique points. Let's take $p,a,b$. Define $T_1=a-p$ and $T_2=b-p$. We can suppose $T_1$ and $T_2$ are orthogonal; if they are not, we can use Gram-Schmidt orthonormalization.
Now, suppose we walk around between $p$ to $a$ or $b$. This is the same as adding some multiple of $T_1$ or $T_2$ to $p$. So a parametric equation for the plane is
$$
x(s,t) = p + sT_1 + tT_2
$$
so that if you input any $s$ and $t$, your output is a point on the plane.
See also here.
Notice that a normal vector is simply $n=(T_1\times T_2)/||T_1\times T_2||_2$.
So we get equivalence to $n\cdot(x-p)=0$ as before.
