A detailed basic-level explanation of equations of lines and planes in 3-d geometry I've searched multiple blogs but couldn't find anything helpful for my level. (I'm in the 12th grade learning about vectors in maths). 
I basically need some thorough explanations regarding plane and line equations in 3 dimensions. I have a lot of unanswered questions which i'm hoping someone will answer!
PS: Please don't use any high-level mathematical notation or anything; I get very confused as i'm a newbie to the entire concept. 
So here's what I know/don't know: (My questions are in bold.)
I know that the equation of a line in 2-d is something of the form ax + by = c. (1. What is this form of a line called?)
Similarly, I know that the equation of a plane is of the form ax + by + cz = d. I know that for all x-, y- and z-coordinates of a point on the plane, ax + by +cz will give d. However, I was also told that planes are referred to in terms of their normals. So how to express a plane in terms of the normal to the plane?
These are all the questions that come to mind for now; but I may end up having more.
Thanks in advance for being patient and explaining to a beginner!
 A: $ ax + by = c $ or equivalently $ ax + by -c = 0 $ are usually referred to as an implicit equation or general (or standard) form for a line in $ \mathbb R^2 $ (the corresponding explicit equation or slope-intercept form would be either $ y =  -\frac{a}{b}x + \frac{c}{b} $, if $ b \ne 0 $, or $ x = \frac{c}{a} $ in case $ b = 0 $, that is, for a vertical line).
A line may be regarded as the locus of points satisfying the above linear equation in 2 dimensions. Likewise, the locus of points satisfying the linear equation in 3 dimensions $ ax + by + cz = d $ is a plane in $ \mathbb R^3 $.
If you have a line in $ \mathbb R^3 $, there is a whole family of planes having that line as a normal (just shift the planes along the direction of the line without changing their orientation in space). 
Up to a translation, we may therefore restrict our attention to planes passing through the origin and normal lines passing through the origin.
In this case, a point whose Cartesian coordinates are $ (x\ y\ z) $ will belong to the plane if and only if
$$ ax + by + cz = 0 $$
(with $ d = 0 $). This may be read as the ordinary scalar product between vectors $ (x\ y\ z) $ and $ (a\ b\ c) $; two vectors whose scalar product is zero are perpendicular. Therefore, the point whose coordinates are $ (a\ b\ c) $ must lie by definition on the normal to the plane. Since the line is through the origin, any other point on that line is found by multiplying $ (a\ b\ c) $ by any real number.
Vice-versa, if you have a line whose direction is specified by the vector $ (a\ b\ c) $, the plane through the origin having that line as a normal is specified by the equation $ ax + by + cz = 0 $.
