(1+2m)x+(1-m)y-3-3m=0
x+2mx+y-my-3-3m=0
x+y+(2x-y-3)m-3=0
恒过定点,则与m的取值无关,所以可得含m项的系数为0,可得：
2x-y-3=0 ····················1
此时原方程可化为：
x+y-3=0·····················2
联立1、2两式得：
x=2,y=1
综上可得直线恒过定点(2,1).