$ax + by + cz + d = 0$ - How does $D$ displace the plane? If $ax + by + cz = 0$ defines the orientation of a plane which passes through the origin, how is it that adding the constant $d$, the position of the plane moves?
I understand that $d$ is the distance from the origin, however it doesn't indicate which direction. Suppose you want to move the plane up by $5$. How does the result differ from wanting to move forward by $5$? In both cases, $5$ is added.
 A: We can get the nice normal form like this:
$$
a x + b y + c z = d \iff \\
(a, b, c) \cdot (x,y,z) = d
$$
Now if $n = (a, b, c)$ is not zero then we can divide by the norm and get
$$
\frac{n}{\lVert n\rVert} \cdot (x,y,z) = \frac{d}{\lVert n \rVert} = D
$$
Now the right hand side $D$ is the (possibly signed) distance to the origin of the plane with unit normal vector $n/\lVert n \rVert$.
Let $u_r = r e_r$ be the vector from the origin to the closest point on the plane, where $e_r$ is a unit vector. Then we have
$$
D = \frac{n}{\lVert n\rVert} \cdot r e_r = r (\pm 1)
$$
depending on what $n$ (we have two sides) was chosen.
So $D$ ist the distance up to a sign.
A: Given a plane
$$
a x + b y + c z + d = 0
$$
Assuming that $a^2+b^2+c^2 > 0$ there exist $x_0,y_0,z_0$ such that
$$
a(x-x_0)+b(y-y_0)+c(z-z_0) = a x + b y + c z + d = 0
$$
by choosing $x_0,y_0,z_0$ such that
$d = a x_0+b y_0+c z_0$
so the existence of $d \ne 0$ implies on an origin translation on the original plane
$$
a x+b y+c z=0
$$
to a point $p_0 = (x_0,y_0,z_0)$
The feeling of translation should be obtained by adopting a plane orientation. This orientation can be associated to the plane normal vector orientation  $\vec n = (a,b,c)$ so we can feel the translation sense by analyzing the sign of $< p_0, \vec n >$
