Using determinants, find the value of k so that

If the points $(k, 2-2 k),(-k+1,2 k)$ and $(-4-k, 6-2 k)$ are collinear, then

$\Delta=\left|\begin{array}{ccc}k & 2-2 k & 1 \\ -k+1 & 2 k & 1 \\ -4-k & 6-2 k & 1\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{ccc}k & 2-2 k & 1 \\ -2 k+1 & 4 k-2 & 0 \\ -4-k & 6-2 k & 1\end{array}\right|=0 \quad$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ ]

$\Rightarrow\left|\begin{array}{ccc}k & 2-2 k & 1 \\ -2 k+1 & 4 k-2 & 0 \\ -4-2 k & 4 & 0\end{array}\right|=0 \quad$ [Applying $R_{3} \rightarrow R_{3}-R_{1}$ ]

$\Rightarrow\left|\begin{array}{cc}-2 k+1 & 4 k-2 \\ -4-2 k & 4\end{array}\right|=0$

$\Rightarrow-8 k+4+16 k-8+8 k^{2}-4 k=0$

$\Rightarrow 8 k^{2}+4 k-4=0$

$\Rightarrow(8 k-4)(k+1)=0$

$\Rightarrow k=-1$ or $k=\frac{1}{2}$