The three given points are A(7, −2), B(5, 1) and C(3, 2k). It is also said that they are collinear and hence the area enclosed by them should be 0.

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

∴ 0 = |7(1 – 2k) + 5(2k – (-2)) + 3(-2 – 1)|

∴ 0 = |7 – 14k + 10k +10 -6 -3|

∴ - |8 – 4k|= 0

8 – 4k = 0

-4k = -8

∴ k = 2