For what value of k will the following pair of linear equations have infinite solutions?

$k x+3 y-(k-3)=0$

$12 x+k y-k=0$

6 Only

$k x+3 y-(k-3)=0$

$12 x+k y-k=0$

a1 = k, b1 = 3, c1 = -(k-3)

a2 = 12, b2 = k and c2 = -k

For infinite solutions, a1/a2 = b1/b2 = c1/c2

k/12 = 3/k = -(k-3)/-k

$k^2 = 36$ ⇒ k = ± 6

and k-3 = 3 ⇒ k = 6 

The correct answer is Option (2) → 6 Only