understanding constants in standard plane equation Currently I am learning about linear algebra and planes, but most tutorials on "how to find the plane equation given a dot and normal vector" assume that I already understand the equation $$Ax + By + Cz = D$$ 
I've never seen this formula before, and I don't understand how this equation represents a plane. From the Internet I understand that $A, B, C$ are components of the normal vector. I also believe this equation is somewhat similar to a line in $R^2$, which is $$Ax + By = C$$
For this second equation I can solve for $y$ and plot the formula. Plotting a 3D plane doesn't strike me as easy though. This stops me from building up intuition for this equation: Changing the values of $m$ or $b$ in $y = ax + b$ changes the slope and the intercept of the line. Conversely, I don't understand the behaviour of each constant in the plane equation. 
What effect will changing the value of $A$ have on the plane equation? And why does the plane equation use the normal vector's components as constants? 
 A: First of all, if don't have some dedicated software (such as Mathematica or Maple) you can plot your plane here (and you right - fiddling with the constants is a great way to build up intuition!).
All right, so the first thing to note is that the equation for the plane is linear in its variables, i.e., no single variable is taken to some power, or put into a trigonometric function, etc., it's just there with its constant... This is what makes the plane flat!
A constant (e.g. $A$) associated with a variable (e.g. $x$) indicates how much the value of the dependent variable (for instance $z$) grows when you change the variable ($x$), just like a normal 2D function $y=Ax$, i.e., $A$ is the slope for that direction. 
And as is the case with $B$ in the case of $y=Ax+B$, the constant $D$ simply raises (if it is positive) the entire plane. 
I hope that helps. If you have any doubts, let me know in the comments. 
A: A very easy intuition  of a plane equation is with 2 things:
1) A point on the plane, say $P_0=(x_0 , y_0 , z_0) $
2) The normal vector, perpendicular to each line on plane surface, say $N=<A,B,C>$
Then for any point on the plane $P=(x,y,z)$, $|PP_0|$ vector is on the surface of the plane, so it is perpendicular to the normal vector. That is the dot product is 0.
$|PP_0| = (x-x_0, y-y_0, z-z_0)$
$|PP_0|.N = A*(x-x_0) + B*(y-y_0) + C*(z-z_0) = 0$
Here you have the plane equation $Ax + By + Cz + D = 0$
A: $(D/A,0,0)$ is the intersection of the plane with $x$-axis, so if you change $A$ you change this intersection and simulanously hould intersectionpoints with $y$- and $z$-axis, just the same for a line in $\mathbb{R}^2$.
You can imagine a plane as the set of vectors, with the same projektion on some normal vector $(A, B, C)$. Projection is given by scalar product, so you get
$$Ax+By+Cz=\text{const.}$$
