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Suppose we have the analytic equation of a plane $(p):Ax+By+Γz+Δ=0$

I know that if none or one exactly of $\{Α,Β,Γ\}$ is $0$ then equation $(p)$ still represents a plane in $\Bbb R^3$.

But what happens if two or three of $\{Α,Β,Γ\}$ are $0$?

Do we still have a plane or is it the whole vector space $\Bbb R^3$?

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    If exactly two coefficients are zero, the remaining equation reads $\text{variable}=-\frac{\Delta}{\text{coefficient}}$, which describes a plane. If all three are zero, the equation is $\Delta=0$ and doesn't describe anything, except maybe a contradiction if $\Delta\neq 0$. – 2017-02-23

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I know that if none or one exactly of $\{Α,Β,Γ\}$ is $0$ then equation $(p)$ still represents a plane in $\Bbb R^3$.

Indeed and the plane will have a specific orientation: if e.g. $Γ=0$, then the plane will be parallel to the $z$-axis; mutatis mutandis for the other coefficients and corresponding axes.

But what happens if two or three of $\{Α,Β,Γ\}$ are $0$?

Do we still have a plane or is it the whole vector space $\Bbb R^3$?

If two of the coefficients are zero, you still have a plane but again with a very specific orientation: if e.g. $B=Γ=0$, then the plane can be written in the form $x=\alpha$ and it will be parallel to the $yz$-plane; mutatis mutandis for combinations of the other coefficients and the corresponding coordinate planes.

If all three coefficients $A$, $B$ and $Γ$ in $Ax+By+Γz+Δ=0$ are $0$, then you have two cases:

  • if $Δ=0$, the equation is trivially satisfied for all $(x,y,z) \in \mathbb{R^3}$, the solution set is all of $\mathbb{R^3}$;
  • if $Δ \ne 0$, the equation is clearly a contradiction and cannot be satisfied by any $(x,y,z) \in \mathbb{R^3}$; the solution set is empty.
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Recall that how we derive a Cartesian equation for a 3-dimensional plane. We are given a plane in $\mathbb{R}^{3}$, a point $(x_{0},y_{0},z_{0})$ on it, and a normal vector $(a,b,c)$ of the plane (which presupposes $a,b,c$ are not all zero.). Then any vector on the plane takes the form $(x-x_{0}, y-y_{0}, z-z_{0})$ and is perpendicular to $(a,b,c)$. So $$ (a,b,c)\cdot (x-x_{0},y-y_{0},z-z_{0}) = ax + by + cz + (-ax_{0} - by_{0} - cz_{0}) = 0. $$ Let $d:= -ax_{0} - by_{0}-cz_{0}$. Then we get a Cartesian equation $ax+by+cz + d = 0$ for the plane under consideration.

Geometrically, if exactly two of $a,b,c$ are $=0$, say $a=b=0$ without loss of generality, then we have $cz +d = 0$, i.e. $z = -d/c$, so the plane is just a shift of the xy-plane along with the $z$-axis.

The derivation above does not allow of $a=b=c=0$.

However, though logically one can argue that $(0,0,0)$ can be a normal vector of any plane, this is indeed only a trivial case of interest to be ruled out. If you want, you can view $(0,0,0)$ as a degenerate 3-dimensional plane.