If a→=xi^+yj^+zk^,b→=yi^+zj^+xk^ and c→=zi^+xj^+yk^, then a→×(b→×c→) is

a→×(b→×c→)=(a→⋅c→)b→−(a→⋅b→)c→=(yz+yx+zx){(y−z)i+(z−x)j^+(x−y)k^}^Clearly this vector is parallel to (y−z)i^+(z−x)j^+(x−y)k^It is orthogonal to i^+j^+k^ as (y−z)(1)+(z−x)(1)+(x−y)(1)=0It is orthogonal to (y+z)i^+(z+x)j^+(x+y)k^as ,(y−z)(y+z)+(z−x)(z+x)+(x−y)(x+y)=y2−z2+z2−x2+x2−y2=0Also it is orthogonal to  xi^+yj^+zk^