(a+b)×(a-b)=(a+b)×a-(a+b)×b = a×a+b×a-a×b-b×b = 2(b×a)

|(a+b)×(a-b)| = 2|b×a|=24

2) a={ 1，0，0 }； b={0， 1，-2 }； c={2，-2， 1}

a×b={0，2， 1}

Because α, a, b*** plane, α is perpendicular to the normal vector n of this plane, that is, n = a× b.

And α ⊥ C.

Let α={x, y, z}

α c =2x-2y-z=0

α (a×b) =2y+z=0

The solution is z =-2y；; x=2y

That is, α={-2y, y, 2y} and unitized α = {-2/3, 1/3, 2/3}.