Let i, j and k be the unit vectors along A, B and C.
Now Cx(AxB)= Ck x (ABsin90°)k =(ABC) k x k =0 ( Since k x k =0)
Let us see the converse, given (AxB)xC=0, if θ be the angle between A and B and k unit vector perpendicular to both A an B
=> ABsinθ k x C = 0
=> ABsinθ.1.Csinß.u =0 (ß is the angle between k and C and u the unit vector along the perpendicular to both k and C)
=> This condition gives either θ =0 or ß=0. So they are not mutually perpendicular and the converse is not true.
