Given points are A( k+1, 2k), B(3k, 2k+3), and C(5k-1, 5k)

These points will be collinear, if area of the triangle formed by them is zero.

We have,

i.e. 

$$\begin{gathered} \left[\left(\mathrm{k}+1\right) \mathrm{x} \left(2\mathrm{k}+3\right) + 3\mathrm{k} \mathrm{x} 2\mathrm{k} + \left(5\mathrm{k} - 1 \right) \mathrm{x} 2\mathrm{k}\right] - \left[3\mathrm{k} \mathrm{x} 2\mathrm{k} + \left(5\mathrm{k}-1\right) \mathrm{x} \left(2\mathrm{k}+3\right) + \left(\mathrm{k}+1\right) \mathrm{x} 5\mathrm{k}\right] = 0 \\ \Rightarrow 2{\mathrm{k}}^2 + 5\mathrm{k} + 3 + 15{\mathrm{k}}^2 + 10{\mathrm{k}}^2 - 2\mathrm{k} -[ 6{\mathrm{k}}^2 + 10{\mathrm{k}}^2 + 13\mathrm{k} - 3 + 5{\mathrm{k}}^2 + 5\mathrm{k} ] = 0 \\ \Rightarrow 27{\mathrm{k}}^2 + 3\mathrm{k} + 3 -[ 21{\mathrm{k}}^2 + 18\mathrm{k} - 3 ]= 0 \\ \Rightarrow 27 {\mathrm{k}}^2 + 3\mathrm{k} + 3 - 21{\mathrm{k}}^2 - 18\mathrm{k} + 3 = 0 \\ \Rightarrow 6{\mathrm{k}}^2 - 15\mathrm{k} + 6 = 0 \\ \Rightarrow 2{\mathrm{k}}^2 - 5\mathrm{k} + 2 = 0 \\ \Rightarrow 2{\mathrm{k}}^2 - 4\mathrm{k} -\mathrm{k} + 2 = 0 \\ \Rightarrow (\mathrm{k}-2) (2\mathrm{k}-1) = 0 \\ \Rightarrow \mathrm{k}-2 = 0 \mathrm{or} 2\mathrm{k}-1 = 0 \\ \Rightarrow \mathrm{k} = 2 \mathrm{or} \mathrm{k} = \frac{1}{2} \end{gathered}$$