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