If $\left(k,2-2k\right),\left(-k+1,2k\right)$  and $\left(-4-k,6-2k\right)$  are collinear, then the possible value of $k$  is

If $\Delta =0$

$\Rightarrow \frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|=0$

$\Rightarrow \frac{1}{2}\left|k\left(2k-6+2k\right)+\left(-k+1\right)\left(6-2k-2+2k\right)+\left(-4-k\right)\left(2-2k-2k\right)\right|=0$

$\Rightarrow \frac{1}{2}\left|k\left(4k-6\right)+\left(-k+1\right)\left(4\right)+\left(-4-k\right)\left(2-4k\right)\right|=0$

$\Rightarrow \frac{1}{2}\left|4k^2-6k-4k+4-8+16k-2k+4k^2\right|=0$

$\Rightarrow \frac{1}{2}\left|8k^2+4k-4\right|=0$

$8k^2+4k-4=0$

$2k^2+k-1=0$

$k=\frac{1}{2};k=-1$