Answer
The points are colinear.
Work Step by Step
$\left|\begin{array}{lll}
a_{1} & b_{1} & c_{1}\\
a_{2} & b_{2} & c_{2}\\
a_{3} & b_{3} & c_{3}
\end{array}\right|$ = see p.645...
$=a_{1}b_{2}c_{3}+b_{1}c_{2}a_{3}+c_{1}a_{2}b_{3}-a_{3}b_{2}c_{1}-b_{3}c_{2}a_{1}-c_{3}a_{2}b_{1}$
---------------------
Using the given formula with
$(x_{1},y_{1})=(-4,-6)$
$(x_{2},y_{2})=(1,0)$
$(x_{3},y_{3})=(11,12)$
$\left|\begin{array}{lll}
-4 & -6 & 1\\
1 & 0 & 1\\
11 & 12 & 1
\end{array}\right|=$
$=0+(-66)+12-0-(-48)-(-6)$
$=-66+66$
$=0$
The points are colinear