collinear and coplanar points on sphere let us consider following project from calculus book :

now  my afford and several notes to  first part, according to that link
http://answers.yahoo.com/question/index?qid=20110518062027AAPmUyF
points are coplanar  if  determinant created in this link is zero,so let us take  example in  our project and check by matrix
A=[1 -1 2 1;2 -1 4 1;-1 -1 -1 1;1 4 1 1]

A =

     1    -1     2     1
     2    -1     4     1
    -1    -1    -1     1
     1     4     1     1

then determinant of this matrix is  det(A)
ans =
 5

which means that they are coplanar,also for answering question to collinearity,we should check following link
http://mathworld.wolfram.com/Collinear.html
clearly what  non coplanar fact means is that they are independent from each other,that why they are  non collinearity,is it right that assumption?what about gaussian elimination?what should be  coefficient matrix in this case?these points seperately denote  denote  $x$,$y$ and $z$ right?in this case we simple would have  for example for first point
$1+1+4+u-v+2*w=k$
$4+1+16+2*u-v+4*w=k$
$1+1+1-u-v-w=k$
$1+16+1+u+4*v+w=k$
or we get
$6+u-v+2*w=21+2*u-v+4*w=3-u-v-w=18+u+4*v+w$
where to use gaussian elimination?please help me,even if i will leave  
$1+1+4+u-v+2*w=k$
$4+1+16+2*u-v+4*w=k$
$1+1+1-u-v-w=k$
$1+16+1+u+4*v+w=k$
these equations, ,then we would have
$6+u-v+2*w=k$
$21+2*u-v+4*w=k$
$3-u-v-w=k$
$18+u+4*v+w=k$
then clearly we have coefficient matrix,and should we apply  gaussian method there?because $k$ is unknow,for different  $k$,we would have different answers right?
 A: It is easier to decide collinearity/coplanarity of vectors than of points. A set of $n$ points $P_1,\dots,P_n$ is collinear/coplanar if and only if the vectors $P_2-P_1,P_3-P_1,\dots, P_n-P_1$ and collinear/coplanar. One advantage is already seen: we have $n-1$ vectors to deal with. 
The dimension of the linear span of a system of vectors is the rank of the matrix composed of those vectors. If the rank is $1$, the vectors are collinear. If it is $2$, they are coplanar but not collinear. If it is $3$, they are not coplanar. Note that one computation of matrix rank answers both collinearity and coplanarity questions at once.
For example, if given points are 
$$(-1, -1, -1), (1, -1, 2), (2, -1, 4), (1, 4 ,1)$$
we can consider the vectors 
$$\langle 2,0,3\rangle, \langle 3,0,5\rangle, \langle2,5,2\rangle$$ 
formed as above. Put them into a matrix, either as rows or as columns: 
$$\begin{pmatrix} 2&0&3 \\ 3&0&5\\ 2&5&2\end{pmatrix}$$ 
The rank is $3$, thus the vectors are not coplanar, and neither are the points. 
No $4\times 4$ determinants (as in Yahoo! Answer) are needed. 

Note that my answer addresses your question about determining the collinearity of coplanarity of points. It does not address the problem in your book; in fact I don't even see what your question has to do with the problem in the book. 
