We have,

kx – y – 2 = 0 (i)

6x – 2y – 3 = 0 (ii)

Here, a₁ = k, b₁ = – 1 and c₁ = – 2

a₂ = 6, b₂ = – 2 and c₂ = – 3

We know that, for the system having a unique solution it must have:

∴

Hence, option D is correct