Find the value (s) of k for which the points (3k - 1, k - 2), (k, k - 7) and (k - 1,-k - 2) are collinear.
Let A(3k − 1, k − 2), B(k, k − 7) and C(k − 1, −k − 2) be the given points.
Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)
= 
|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
Given that area of ∆ABC = 0
(3k−1)[(k−7)−(−k−2)]+k[(−k−2)−(k−2)]+(k−1)[(k−2)−(k−7)]=0
(3k−1)(2k−5)+k(−2k)+5(k−1)=0
6k2−17k+5−2k2+5k−5=0
4k2−12k=0
4k(k−3)=0
k=0 or k−3=0
k=0 or k=3
Hence, the value of k is 0 or 3.
23