Find if three points in 3-dimensional space are collinear

Method 1:

Point $A$ and point $B$ ($A \ne B$) determine a line. You can find its equation. See if the coordinates of point C fits the equation. If so, A B and C are colinear, or else, no.

Method 2:

Point $A$, $B$ and $C$ determine two vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. Suppose the latter isn't zero vector, see if there is a constant $\lambda$ that allows $\overrightarrow{AB}=\lambda \overrightarrow{AC}$.

Other properties if $A$, $B$ and $C$ are colinear:

$$\left| \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{\left|\overrightarrow{AB}\right|\cdot\left|\overrightarrow{AC}\right|} \right| = 1$$:\

$$\overrightarrow{AB}\times\overrightarrow{AC} = \overrightarrow{0}$$

Also, two ways to write the equation of a line in 3D:

$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$$

where $(x_0,y_0,z_0)$ is a point on the line and $(a,b,c)$ is the direction vector of the line, provided that $abc\ne 0$.

$$ \begin{align} x&=x_0+at,\\ y&=y_0+bt,\\ z&=z_0+ct. \end{align}$$

All that remains is calculation.

Hint $A,B,C$ are colinear if and only if the largest of the lenghts of $AB, AC, BC$ is equal to the sum of the other two.

We have $\overrightarrow{AB}=(-4,-10,0)$ and $\overrightarrow{AC}=(-2,-12,-1)$ . Therefore cross product of two vectors AB and AC is $\overrightarrow{AB}\times\overrightarrow{AC}=(10, -4, 28)$ . This vector is different from vector $(0,0,0)$. So, the given points are not co-linear.