In $\mathbb{R}^3$, have the points
$$A = (2,0,-1)\\ B = (3,1,-1)\\ C = (4,-2,0)$$
Verify that the points don't belong to the same line without drawing.
From what I gather, a line is defined by one point and one direction arrow.
Technically, given two points, $A,B$, they will always belong to the same line, if the direction arrow is $\vec{AB}$. So it all boils down to determining whether the third point, $C$, belongs to such line.
A valid line, $X$, would be
$$X = L(A,\vec{AB})$$
Where
$$\vec{AB} = B - A = (3,1,-1) - (2,0,-1) = (1,1,0)$$
So $X$ is a line with a point at $(2,0,-1)$ and a direction arrow $(1,1,0)$.
IF the third point, $C$, belongs to this line, we should be able to express it as
$$(4,-2,0) = (2,0,-1) + s \cdot (1,1,0)$$
For some $s \in \mathbb{R}$.
The above equation can be seen in parametrized form:
$$(4,-2,0) = (s + 2, s, -1)$$
And we have an equation system:
$$\begin{cases} s+2 = 4\\ s=-2\\ -1 = 0 \end{cases}$$
However this is clearly inconsistent - therefore $C$ cannot belong to the line $X$.
Not sure if I did it right - but I seem to have proved that $C$ does not belong to the line that is drawn using $A$ and $B$. However, my question is: do I need to prove the other cases? That is, prove that $A$ does not belong to the line drawn using $B$ and $C$, etc...?
Off my head, I'm guessing there is no need to.
You are correct
Didn't check the arithmetic, but the approach is correct.
You don't have to check the other cases as well. Special handling of those might be required if two of your points coincide. So for example if $A=B$, then the “line” spanned by these will be only a point, but for any $C$ the points $A,B,C$ will still lie on a single line.
Rank computation
Since a comment suggested a rank computation to approach this problem, here is a different way to look at it:
\begin{align*} C &= A+s(B-A) \\ (B-A)s &= C-A \\ (1-s)A + sB - C &= 0 \\ (1-s)tA + stB - tC &= 0 \\ aA + bB + cC &= 0 \quad\text{and}\quad a+b+c=0 \end{align*}
So you are looking for a non-trivial (i.e. not $a=b=c=0$) solution to the system of equations
$$ \begin{pmatrix} x_A & x_B & x_C \\ y_A & y_B & y_C \\ z_A & z_B & z_C \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix}= \begin{pmatrix}0\\0\\0\\0\end{pmatrix} $$
If you can combine the columns of the matrix in a non-trivial way to obtain the null vector, that means that the columns of the matrix must be linearily dependent, which in turn means that the matrix has rank less than three.
You can check that by performing a Gaussian elimination, or by computing all the four $3\times3$ determinants you can get from omitting one of the four rows (i.e. the $3\times3$ minors). They will all be zero iff the points are collinear.
One can interpret the columns of the matrix as homogeneous coordinates of the three points. In that projective view, the four minors would form the Plücker coordinates of the plane spanned by these three points, which results in the null vector iff that plane is not uniquely defined because the points are collinear.