We know that, if three points are collinear, then the area of triangle formed by these points is zero.

Since, the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k - 1, 5k) are collinear.

Then, area of ΔABC = 0

1/2 [x₁ (y₂ – y₃) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )] = 0

Here,

x₁ = k + 1,

x₂ = 3k

x₃ = 5k – 1

And

y₁ = 2k

y₂ = 2k + 3

y₃ = 5k

[(k + 1)(2k + 3 - 5k) + 3k(5k – 2k) + (5k – 1)(2k + 3)]

[(k + 1)( - 3k + 3) + 3k(3k) + (5k - 1)(2k – 2k – 3)] = 0

[ - 3k² + 3k - 3k + 3 + 9k² - 15k + 3] = 0

(6k² – 15k + 6) = 0 (multiply by 2)

→ 6k² – 15k + 2 = 0

→ 2k² – 5k + 2 = 0

[Divide by 3]

→ 2k² – 4k – k + 2 = 0

→ 2k (k – 2) – 1(k – 2) = 0

→ (k – 2)(2k – 1) = 0

If k – 2 = 0, then k = 2

If 2k – 1 = 0, then k= 1/2

∴ k = 2, 1/2

Hence, the required values of k are 2 and 1/2.