For (2x3+xk3)12
Tr+1=Cr12(2x3)12−r⋅(xk3)r=Cr12⋅212−r⋅3r⋅x36−3r−kr
For the term to be independent of x,
36−3r−kr=0 ⇒r=k+336
Here k can be 0,1,3,6,9 for r to be a whole number.
But for Cr12⋅212−r⋅3r to be in the form of 28⋅l
Only k=3,6 satisfies the above relation.