The given pair of linear equations is

kx + 3y = k – 3 …(i)

12x + ky = k …(ii)

On comparing the equations (i) and (ii) with ax + by = c = 0,

We get,

a₁ = k, b₁ = 3, c₁ = -(k – 3)

a₂ = 12, b₂ = k, c₂ = – k

Then,

a₁ /a₂ = k/12

b₁ /b₂ = 3/k

c₁ /c₂ = (k-3)/k

For no solution of the pair of linear equations,

a₁/a₂ = b₁/b₂≠ c₁/c₂

k/12 = 3/k ≠ (k-3)/k

Taking first two parts, we get

k/12 = 3/k

k² = 36

k = + 6

Taking last two parts, we get

3/k ≠ (k-3)/k

3k ≠ k(k – 3)

k² – 6k ≠ 0

so, k ≠ 0,6

since k is not 6 it must be -6.

k=-6