Answer
Yes.
Work Step by Step
The given coordinates are
$\Rightarrow (x_1,y_1)=(-4,-6)$
$\Rightarrow (x_2,y_2)=(1,0)$
$\Rightarrow (x_3,y_3)=(11,12)$
The determinant to show the three points is
$\Rightarrow \begin{vmatrix}
x_1& y_1& 1\\
x_2& y_2& 1\\
x_3& y_3& 1
\end{vmatrix}$
Plug all the values.
$\Rightarrow \begin{vmatrix}
-4& -6& 1\\
1& 0& 1\\
11& 12& 1
\end{vmatrix}$
Perform $R_2\rightarrow R_2+R_1(1/4)$ and $R_3\rightarrow R_3+R_1(11/4)$.
$\Rightarrow \begin{vmatrix}
-4& -6& 1\\
1+(-4)(1/4)& 0+(-6)(1/4)& 1+(1)(1/4)\\
11+(-4)(11/4)& 12+(-6)(11/4)& 1+(1)(11/4)
\end{vmatrix}$
Simplify.
$\Rightarrow \begin{vmatrix}
-4& -6& 1\\
0& -3/2& 5/4\\
0& -9/2& 15/4
\end{vmatrix}$
Perform $R_3\rightarrow R_3-3R_2$.
$\Rightarrow \begin{vmatrix}
-4& -6& 1\\
0& -3/2& 5/4\\
0-3(0)& -9/2-3(-3/2)& 15/4-3(5/4)
\end{vmatrix}$
Simplify.
$\Rightarrow \begin{vmatrix}
-4& -6& 1\\
0& -3/2& 5/4\\
0& 0& 0
\end{vmatrix}$
All elements below the diagonal are zero.
Hence, the determinant is the multiplication of the main diagonal elements.
$\Rightarrow (-4)\cdot (-3/2)\cdot (0) $
Simplify.
$\Rightarrow 0 $
The determinant is zero.
Hence, the points are colinear.
