If $(k,2-2k),(-k+1,2k),(-4-k,6-2k),$are collinear, then k =

A (k, 2-2k), B(-k+1,2k), C(-4-k, 6-2K)

Slope of AB = Slope of AC $\Rightarrow \frac{2k-2+2k}{-k+1-k}=\frac{6-2k-2+2k}{-4-k-k}\Rightarrow \frac{4k-2}{1-2k}=\frac{4}{-2(k+2)}$

$\Rightarrow 2(2k-1)(k+2)=-2(1-2k)\Rightarrow (2k-1)(k+2)=-(1-2k)\Rightarrow (2k-1)(k+2-1)=0$

$\Rightarrow k=1/2,-1$