Answer
Collinear.
Work Step by Step
The general form of a matrix of order $ 3 \times 3$ is:
$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h& i \end{bmatrix}=a(ei-fh) -b(di-fg)+c(dh-eg)$
Use the formula for the area of a triangle with the determinant.
$D=det \begin{bmatrix} a & x & 1 \\ b & y & 1 \\ c & z & 1 \end{bmatrix} $
$Area=|\dfrac{1}{2} D|$
Now, $D=det \begin{bmatrix} 0 & -5 & 1 \\ -2 & -6 & 1 \\ 8 & -1 & 1 \end{bmatrix} =0$
Because the determinant is zero, the three points are Collinear.
