Q:  f(L,K)=3L+2K.  In the short run Labor is fixed at L=2.  If w=2 and v=1 what is the cheapest way to make 12 units of output in the short run?

A:  Since L=2 is fixed in the short-run, to make 12 units of output, it must be that f(2,K)=3(2)+2K=12.  Or that 6+2K=12.  This means that 2K=6 or K=3.  Therefore the optimal bundle is L=2 and K=3.  The cost of this is wL+vK=2(2)+1(3)=7. 
You should note that the cheapest way to produce 12 units in the long-run is with 6 units of K and 0 units of L which would only cost 6.