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.
Best Answer
This is how your vector field looks like. Taken from http://www.lightandmatter.com/html_books/lm/ch22/ch22.html
That is, your vector field is the gravity forces field generated by a mass at the origin, up to a constant. To graph it it might help to note that the modulus of $F$ is a radial function and that the direction points to the origin...