To preface the question, with respect to the topic at hand, I am an absolute beginner. I've only very recently dived into a lesson in my book that deals as an intro of sorts to vectors and planes in 3D. Searches in regards to this question on the Internet, if any, only leave me more confused

I've thus far, learned about various ways to represent planes:

1. With the normal vector of the plane + distance from the origin : $\vec a \bullet\hat n = d$
2. With the normal vector of the plane + a point on the plane : $(\vec r - \vec a) \bullet \hat n = 0$
3. With 3 non-collinear points : $(\vec r - \vec a).((\vec b - \vec a) \times (\vec c - \vec a)) = 0$

The fourth one thus far, the intercept form , broaches a new equation for a plane (before moving on to the rest of the derivation) , of which I've not seen before:

$Ax + By + Cz + D = 0$

I'd initially thought that this was a rearrangement of $(\vec r - \vec a) \bullet \hat n = 0$

That is, if $\vec r$ was some vector of the form $x \hat i + y \hat j + z \hat k$, and $\hat n$ was $A \hat i + B \hat j + C\hat k$ , it would yield $A \hat i + B \hat j + C\hat k - D = 0$, since $\vec a \bullet \hat n$ being the distance from the origin to the plane

However, the equation given in the derivation of the intercept form is $Ax + By + Cz + D = 0$, wherein we do not find a negative D

How exactly was $Ax + By + Cz + D = 0$ 'derived', so to speak. I can't seem to make sense of it or intuit it

To preface the question, with respect to the topic at hand, I am an absolute beginner. I've only very recently dived into a lesson in my book that deals as an intro of sorts to vectors and planes in 3D. Searches in regards to this question on the Internet, if any, only leave me more confused

I've thus far, learned about various ways to represent planes:

1. With the normal vector of the plane + distance from the origin : $\vec a \bullet\hat n = d$
2. With the normal vector of the plane + a point on the plane : $(\vec r - \vec a) \bullet \hat n = 0$
3. With 3 non-collinear points : $(\vec r - \vec a).((\vec b - \vec a) \times (\vec c - \vec a)) = 0$

The fourth one thus far, the intercept form , broaches a new equation for a plane (before moving on to the rest of the derivation) , of which I've not seen before:

$Ax + By + Cz + D = 0$

I'd initially thought that this was a rearrangement of $(\vec r - \vec a) \bullet \hat n = 0$

That is, if $\vec r$ was some vector of the form $x \hat i + y \hat j + z \hat k$, and $\hat n$ was $A \hat i + B \hat j + C\hat k$ , it would yield $Ax + By + Cz - D = 0$, since $\vec a \bullet \hat n$ being the distance from the origin to the plane

However, the equation given in the derivation of the intercept form is $Ax + By + Cz + D = 0$, wherein we do not find a negative D

How exactly was $Ax + By + Cz + D = 0$ 'derived', so to speak. I can't seem to make sense of it or intuit it