Three Variable Linear Equations

Question

Why do three variable linear equations (ax+by+cz=d, where a, b, c are constants not equal to zero) outline a plane in rectangular space?

Answer 1

Pick any arbitrary point $(x_0,y_0,z_0)$ which satisfies $$ax_0+by_0+cz_0=d$$

Now if you pick any point $(x,y,z)$ which satisfies $$ax+by+cz=d$$

We have the relation $$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$

That is the vector $<a,b,c>$ is orthogonal to the vector $(x-x_0,y-y_0,z-z_0)$

As a result the point $(x,y,z)$ runs over a plane passing through the point $ (x_0,y_0,z_0)$ and is perpendicular to the vector $<a,b,c>.$

Answer 2

I think I came up with an answer. Obviously a 3 variable linear equation defines a surface where each coordinate (x,y) creates a height "z". The question is then, is this surface bumpy and irregular or flat as in a plane?

Well if we hold "y" constant and look at a cross section of the graph on the x,z coordinate grid, we would have to see a straight line. This is because holding "y" constant creates a two variable linear equation which defines a straight line. This tells us the graph is straight at any cross section when viewed head on from the y-axis.

And if we hold "x' constant, the same thing occurs. Holding "x" constant creates a two variable linear equation which defines a straight line. This means viewing the graph head on from the x-axis, at any cross section, we would see a straight line.

These two bits of information, put together, tell us that a three variable linear equation must define a flat plane in three dimensions.

Is this a good/accurate explanation?