f(x1,x2,x3)=x1^2-3x2^2-2x1x2+2x1x3-6x2x3
= (x1-x2+x3)^2-4x2^2-x3^2-4x2x3
= (x1-x2+x3)^2-(2x2-x3)^2
所以二次型的符号差为1-1=0.