Why do the equations $Ax + By + Cz= D$ represent planes in $\Bbb R^3$ and not lines? I was wondering why $$Ax+By+Cz=D$$ represents a plane in $\mathbb{R}^3$, because from my perspective it seems that it could simply also represent a line in  $\mathbb{R}^3$. And, yes, I do acknowledge that lines aren't three-dimensional, but surely we can have lines in three-dimensional space. 
 A: If you know about dot product, the above equation can be written as (assume $C\neq 0$)
$$ \langle A, B, C\rangle \cdot \langle x, y, z - \frac{D}{C}\rangle = 0$$
Thus $(x, y, z)$ satisifies $Ax+By+Cz = D$ if and only if the vector from $(0,0,\frac{D}{C})$ to $(x, y, z)$ is perpencidular to $\langle A, B, C\rangle$. The set of all such $(x, y, z)$ will be a plane passing through $(0, 0, \frac DC)$ with normal $\langle A, B, C\rangle$. 
A: If you want a line in three dimensions you need two such equations. The reason is that a line is one dimensional object, so you must apply two constraints to reduce the three dimensional space to a one dimensional object. If you only apply a single constraint, you are left with $3-1= 2$ degrees of freedom, leaving you with a two dimensional object, in this case a plane.
In general, in a $n$ dimensional space, you'll need to write down $n-1$ equations to describe a line, $n-2$ to describe a plane, etc.
N.B
This of course has a linear algebra connection: If we have $n-k$ independent linear equations, the rank of the matrix in the resulting matrix equation is $n-k$, and therefore the dimension of the null space must be $k$. In other words, the object described is spanned by $k$ vectors. 
A: how about write it in another form like $Ax + By = D - Cz$, and treat z as a constant. 
If $z = 0$, we get a line, if $z=1$, we get anther line. 
When change $z$ from minus infinity to positive infinity.  We got lots of lines, which makes them a hyperplane. Maybe you can draw them on a paper.
A: Trivially, if $A = B = C = 0$, then the solution set will be all of $\Bbb R^3$ (if $D = 0$) or the empty set (if $D \neq 0$).
So, suppose not, that is, that at least one of $A, B, C$ is nonzero; by relabeling if necessary, $C$ is. Rearranging gives
$$z = -\frac{A}{C} x - \frac{B}{C} y + \frac{D}{C},$$
so the solution set is the graph of an affine function of $x, y$, and hence is a plane.
One can specify a line by a suitable pair of such equations,
$$A_i x + B_i y + C_i z = D_i, \qquad i = 1, 2.$$
(This system defines a line iff (1) at least one of $A_i, B_i, C_i$ is nonzero, for both $i = 1, 2$, and (2) the coefficient vectors ${\bf A}_1 := (A_1, B_1, C_1), {\bf A}_2 := (A_2, B_2, C_2)$ are linearly independent. Condition (2) is equivalent to ${\bf A}_1 \times {\bf A}_2 \neq 0$. Indeed, the line is parallel to that cross product.)
A: I hope you’re not surprised that $z=0$ describes the $(x,y)$-plane: all points $(x,y,0)$ in space; nor that $y=0$ describes the $(x,z)$-plane: all $(x,0,z)$ in space. What line could either of these equations possibly describe?
Actually, there is a way of describing a line in space by a single equation, as long as you’re talking about real points only, but it’s a cheat. The equation $y^2+z^2=0$ describes the $x$-axis, because for real numbers, the only way that a sum of squares can be zero is for the individual summands each to be zero, so this equation boils down to $y^2=0$ together with $z^2=0$, inother words $y=0$ and $z=0$.
A: If we fix a value of x and y, we get only one value of z. i.e x and y can take any real value but then z will be fixed.This is the algebraic way of stating a plane.
