Intuition behind normal vector of a plane In another question here someone asks why a normal vector of a plane is simply a vector of the coefficients. For example:
For the plane $ 2x−y+3z=8 $, a normal vector is $(2,−1,3)$
The result is explained by saying if you take any 2 arbitrary points on the plane, you have $⟨x1−x2,y1−y2,z1−z2⟩⋅⟨2,−1,3⟩=0$ , and thus the coefficients must be the norm to the plane. 
What I'm not understanding is why you can assume the coefficients dotted with the two points on the plane must equal zero. What is the intuition behind that assumption?
 A: The fact is that every vectors in the plane is orthogonal to to the normal vector.
Since for all $(x_i,y_i,z_i)$:
$(x_1−x_2,y_1−y_2,z_1−z_2)$ represent a vector parallel to the plane
and since by the plane equation:
$⟨x_1−x_2,y_1−y_2,z_1−z_2⟩⋅⟨2,−1,3⟩=0$
thus
$n=(2,−1,3)$ is a normal vector to the plane.
A: As normal vectors to a plane are (in some sense) a generalization of the slope of a line, it may be helpful to first think in two dimensions before generalizing to higher dimensions.  In two dimensions, a line can be given by an equation of the general form
$$ \alpha x + \beta y = c
\iff y = -\frac{\alpha}{\beta}x + \frac{\gamma}{\beta}, \tag{1}$$
where $\alpha,\beta,\gamma \in \mathbb{R}$ are constants.  In slope-intercept form, this becomes
$$ y = mx + b,$$
where $m = -\frac{\alpha}{\beta}$ is the slope and $b = \frac{\gamma}{\beta}$ is the $y$-intercept.  We know from elementary algebra that the equation of the line perpendicular to this line through a given point $(x_0, y_0)$ is
$$ y - y_0 = -\frac{1}{m}(x-x_0)
\implies y = -\frac{1}{m}x + b', $$
where $b' = \frac{x_0}{m} + y_0$ is the $y$-intercept of the perpendicular line.  This can be shown by purely geometric means, and doesn't rely on any vector-flavoured ideas.  Notice that we can parameterize this perpendicular line as the set of points
$$ \left\{ t\left\langle 1, -\frac{1}{m} \right\rangle + (0,b') \ \middle|\ t \in \mathbb{R} \right\} $$
From this parameterization, it is hopefully clear that the vector $\left\langle 1, -\frac{1}{m} \right\rangle$ is normal to the original line (since the parameterized line is in the direction of this vector, and perpendicular to the original line).  Recalling the definition of $m$, this becomes
$$ \left\langle 1, -\frac{1}{m} \right\rangle = \left\langle 1, \frac{\beta}{\alpha} \right\rangle. $$
Finally, since multiplication by a scalar does not change the direction of a vector (up to a reflection, which is not relevant in this situation), we can multiply this vector by $\alpha$ in order to obtain a vector
$$ \langle \alpha, \beta \rangle, $$
which is also orthogonal to the original line given in (1).  But observe that this is exactly the normal vector that we would have obtained using the formula given in the original question!

Moving up to three dimensions, it is reasonable to suspect that a similar formula will hold, i.e. if a plane is given by
$$ \alpha x + \beta y + \gamma z = \delta, \tag{2}$$
then the vector $\langle \alpha, \beta, \gamma \rangle$ will be normal to the plane.  However, the geometric intuition is a little more difficult, so we might prefer to work through this problem using more vectory-flavored ideas.  We want to show that if $\langle a, b, c \rangle$ is any vector in the plane, then we must have
$$ \langle a,b,c\rangle \perp \langle \alpha,\beta,\gamma \rangle.$$
Recall that these two vectors are orthogonal if and only if
$$ \langle a,b,c, \rangle \cdot \langle \alpha,\beta,\gamma \rangle = 0, \tag{3} $$
hence we want to show that (3) holds whenever $\langle a,b,c \rangle$ is a vector in our specified plane.  So we need to get our hands on a general vector in this plane.  One way that we can do this is by considering any two points in the plane, and taking our typical vector to be the arrow between them.  That is, if
$$ (x_1,y_1,z_1) \qquad\text{and}\qquad (x_2,y_2,z_2) $$
are both points in the plane specified at (2), then the vector
$$ \langle x_1-x_2, y_1-y_2, z_1-z_2 \rangle $$
must be in the plane (or, really, parallel to the plane, which is all that really matters).  But then
\begin{align}
\langle x_1-x_2, y_1-y_2, z_1-z_2 \rangle \cdot \langle \alpha,\beta,\gamma \rangle
&= \alpha(x_1 - x_2) + \beta(y_1 - y_2) + \gamma(z_1-z_2) \\
&= \left( \alpha x_1 + \beta y_1 + \gamma z_1 \right) + \left( \alpha x_2 + \beta y_2 + \gamma z_2 \right) \\
&= \delta - \delta \tag{4}\\
&= 0
\end{align}
where equality at (4) follows from the fact that the two points specified at (3) are in the plane and must therefore satisfy (2).  In particular, this shows that if a plane is specified by the equation
$$ \alpha x + \beta y + \gamma z = \delta, $$
then the vector
$$ \langle \alpha, \beta, \gamma \rangle $$
will be normal to the plane.
