Consider the vectors, a=2i−3j+kb=i−j+2k And c=2i+j+k So (a×b)×c=(c⋅a)b−(c⋅b)a=(2i−2j+4k)−(6i−9j+3k)=−4i+7j+k=66 And, a×(b×c)=(a⋅c)b−(a⋅b)c=(2i−2j+4k)−(14i+7j+7k)=−12i−9j−3k=234 Therefore, ∣a×(b×c)∣∣(a×b)×c∣=23466(a×b)×c=3911a×(b×c)