Find value of k for which …

Given points are A(3k -1, k - 2), B(k, k - 7) and C(k-1, -k - 2)
We know that points A, B, C will be collinear, if the area of the ΔABC =0
Area of ΔABC={tex}\frac{1}{2}{/tex}|x₁(y₂ - y₃)+x₂(y₃ - y₁)+x₃(y₁ - y₂)|
here, x₁ =3k-1, x₂ = k, x₃= k-1, y₁= k-2, y₂= k-7, y₃= -k-2
Area of ΔABC = 0
{tex}\Rightarrow{/tex} {tex}\frac{1}{2}{/tex}|(3k−1)[(k−7)−(−k−2)]+k[(−k−2)−(k−2)]+(k−1)[(k−2)−(k−7)]|=0
{tex}\Rightarrow{/tex} {tex}\frac{1}{2}{/tex}|(3k−1)(k−7+k+2)+k(−k−2−k+2)+(k−1)(k−2−k+7)|=0
{tex}\Rightarrow{/tex} |(3k−1)(2k−5)+k(−2k)+(k−1)(5)|=0
{tex}\Rightarrow{/tex} |3k(2k−5)−1(2k−5)−2k²+5k−5|=0
{tex}\Rightarrow{/tex} |6k²−15 k−2k+5−2k²+5k−5|=0
{tex}\Rightarrow{/tex} |4k² - 10k - 2k|=0
{tex}\Rightarrow{/tex} 4k² - 12k = 0
{tex}\Rightarrow{/tex} 4k(k-3) = 0
{tex}\Rightarrow{/tex} k = 0 or k - 3 =0
{tex}\Rightarrow{/tex} k = 0 or k = 3