How do 3 points define a plane? I was solving a combinatorics problem which asked me to find the number of planes that can be constructed from a set of 25 points such that no 4 points in the set of 25 points are co-planar and then I researched about the questions as to how many points define a plane in general and came across the following link - https://answers.yahoo.com/question/index?qid=20090930201928AAL3FeN which says 3 points are sufficient to define a plane. Can someone try to explain more generally using some examples as to how 3 points define a plane?
 A: If you look at the vector equation of a plane

$$P:\; \mathbf{p}+t\mathbf{u}+s\mathbf{v}$$

it's simple:  you need two vectors, $\mathbf{u},\mathbf{v}$, (linearly independent) and a point, $\mathbf{p}$. Given three points, we can take any two of them (assuming they're non collinear, which of course is necessary) and write out their coordinate differences to form vectors which we can use to write the plane. So what's the plane passing through say the points $(0,0,0), (1,0,0), (0,1,0)$?
Well, let's use the point $\mathbf{p}=(0,0,0)$ as the base point and form our direction vectors $\mathbf{u}=(1,0,0)-(0,0,0)=(1,0,0)$ and $\mathbf{v}=(1,0,0)-(0,1,0)=(1,-1,0)$ (you can pick any two of them, I chose these two for ease of computation and to illustrate the computation where both vectors don't end up looking exactly the same as the point) This gives us a plane equation:

$$P:\; t\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+s\begin{pmatrix} 1 \\ -1 \\ 0\end{pmatrix}=\{(x,y,z)\in\Bbb R^3 : x=t+s,\, y= -s,\, z=0\}$$

In more dimensions you do the same procedure, only with more coordinates. Let me demonstrate in $\Bbb R^4$ with points $(1,0,0,0),(3,0,0,0), (0,1,0,0)$ then I'll choose my base point to be $\mathbf{p}=(1,0,0,0)$ and my two direction vectors $\mathbf{u}=(3,0,0,0)-(1,0,0,0)=(2,0,0,0)$ and $\mathbf{v}=(3,0,0,0)-(0,1,0,0)=(3,-1,0,0)$ giving us
$$P:\; \begin{pmatrix} 1 \\ 0 \\ 0 \\0\end{pmatrix}+t\begin{pmatrix} 2 \\ 0 \\ 0 \\0\end{pmatrix}+s\begin{pmatrix} 3 \\-1\\ 0 \\ 0 \end{pmatrix}$$
$$=\{(x,y,z,w)\in\Bbb R^4: x=1+2t+3s,\, y= 2t-s,\, z=w=0\}.$$
Now there's a problem with $4$ points, namely the same problem with why $3$ points don't always determine a line:  if, for example, you asked what line are the points $(0,0,0),\, (1,0,0), \, (0,1,0)$ was in the first question you'd laugh at me. Why? Because there is no such line:  necessarily all points on a line have to have the same direction vector, since--by definition--a line is a set of points in $\Bbb R^n$ such that:

$$\ell:\; \mathbf{p}+t\mathbf{v}$$

for some base point, $\mathbf{p}$ and direction vector, $\mathbf{v}$.
When you have three points, and try to make all of them fit that equation, you get a system of equations that might not be solvable because there are too many restrictions, just like asking for positive numbers less than $-2$, i.e

$$\{x\in\Bbb R : x>0\text{ and } x<-2\}$$

You can ask for positive numbers (there are a TON) and you can ask for those less than $-2$, but just because you have two conditions doesn't mean they're compatible, it sometimes ends up that you some of them try and fight the others. The same is true with $4$ points:  since $3$ points determine the plane, it can be that you can pick another point in space which is not on that plane (unless you're working in $\Bbb R^2$ this is always possible). In which case, if you move your plane to include the missing points, you've got to move it away from the three you've already set up. Now, as before you can always keep three of the four, but there's no guarantee you can get them all:  most times it just isn't in the cards.
A: With 3 points defining a plane, the number of combinations is $^{25}C_3.   $ 
A surface going through 4 points has Gauss curvature positive or negative, never zero unless the fourth point is in the plane determined by the first three.
