Is making vectors a good way to determine the coplanarity of 4 points? I'm given this task:

Explain how you would prove if four given points are coplanar. Use your method to determine if A (5, 1, 7), B (13, -5, 17), C (-1, 15, 3) and D (15, 5, 23) are coplanar.

I'm wondering if the method I'm thinking of would work: I'm assuming that if these 4 points are coplanar, then vectors drawn between them would be coplanar as well. So if I were to make the vectors $\vec {AB}$, $\vec {AC}$ and $\vec {AD}$, then determining their coplanarity would also determine the coplanarity of the 4 points? That is, if those vectors are coplanar, then so are the points, if they aren't, neither are the points.
Or am I going about it wrong? I tried doing this, but my answer had some odd values on the right hand side. It looked unusally "dirty" to be a school-manufactured problem, because they usually have very neat-looking solutions.
I ended up with this equation when trying to find their coplanarity: $$[8,-6,10]=[-6s,14s,-5s]+[10t,4t,16t]$$
From there I found through the equations given by singling out the x and y components that the scalars had to be something along the lines of $t=\frac {80}{174}$ and $s=\frac {19}{29}$ or something. Obviously the third equation did not work out when I substituted in those values (not those exact ones, I don't remember the exact ones. They were similar, though).
Am I doing it wrong, or did I just miscalculate somewhere? Or did I do it right?
 A: One way to do this would be:
Take the cross product of any of any 2 vectors(say $\vec {AB}$, $\vec {AC}$) . Then take the dot product of the resulting vector with the remaining one(here $\vec {AD}$). If it is 0,then  $\vec {AB}$, $\vec {AC}$, $\vec {AD}$ will be co-planar.
A: Yes, your way works, but you need to pay attention to a few points.
What you are trying to do is to check if $B$ is on a plane where three points $A,C,D$ exist using
$$\vec{AB}=s\vec{AC}+t\vec{AD}.$$
(By the way, beforehand, you need to check there is no $m\in\mathbb R$ such that $\vec{AC}=m\vec{AD}.$ Otherwise, three points $A,C,D$ may be on a line. Fortunately, we can easily know that the three points are not on a line.)
Hence, we know that there is such $(s,t)$ if and only if $B$ is on the plane. 
However note that 
$$[8,-6,10]=[-6s,14s,-5s]+[10t,4t,16t]$$
has to be 
$$[8,-6,10]=[-6s,14s,-4s]+[10t,4t,16t].$$
In this case, there is no such $(s,t)$. So this tells you...
A: Basically it is the correct idea to form vectors and check their coplanarity. However you seem to build a plane with 3 points and check wether the 4th point is on it (which would work, too). 
The standard way would be to check the coponarity of the vectors (which means: check whether they are linarily independent)
