a•b= b•c=c•a=1*1*cos120°=-1/2.
|ka+b+c|=√[(ka+b+c)•(ka+b+c)]=
√(k^2|a|^2+|b|^2+|c|^2+2ka•b+2b•c+2kc•a)
=√(k^2+1+1-k-1-k)
=√(k^2-2k+1)=|k-1|,
所以|k-1|>1,
解得k>2或k<0