Scalar versus linear equation of a plane What is the difference between a scalar and a linear equation of a plane? In my textbook it says that a scalar equation is $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
and a linear equation is $ax+by+cz=d$
How do they differ in terms of what they describe?
If a line on the plane dot the normal vector $= 0$ why do we have $d$ for the linear equation?

The two equations describe precisely the same sets. Suppose that a non-zero vector $n = (a, b, c)$ and a point $p_{1} = (x_{1}, y_{1}, z_{1})$ are given, and let $p = (x, y, z)$ denote an arbitrary point of $\mathbf{R}^{3}$.
Expanding the "scalar" equation gives
\begin{align*}
0 &= a(x - x_{1}) + b(y - y_{1}) + c(z - z_{1}) \\
  &= ax - ax_{1} + by - by_{1} + cz - cz_{1} \\
  &= ax + by + cz - \underbrace{(ax_{1} + by_{1} + cz_{1})}_{\text{Call this $d$}},
\end{align*}
or $ax + by + cz = d$, what you call the "linear equation".
There is no unique way to go "backward" from the linear equation to a scalar equation: If $(x_{1}, y_{1}, z_{1})$ lies on the plane with "linear" equation $ax + by + cz = d$, namely if $ax_{1} + by_{1} + cz_{1} = d$, then
$$
a(x - x_{1}) + b(y - y_{1}) + c(z - z_{1}) = 0.
$$
Each can be written as a vector equation:
$$
0 = n \cdot (p - p_{1})\quad\text{("scalar" form)};\qquad
n \cdot p = n \cdot p_{1}\quad \text{("linear" form)}.
$$
The "scalar" equation is natural to write down when $n = (a, b, c)$ and a point $p_{1} = (x_{1}, y_{1}, z_{1})$ are known. The "linear" equation is sometimes a bit easier to use in computations with specific planes.
Neither equation is more correct than the other; they're both descriptions of the plane in space through the point $p_{1}$ and with normal vector $n$.
