In scalar triple product, the position of dot and cross can be changed provided the cyclic order is maintained i.e., [abc]=(a×b)⋅c=a⋅(b×c)
Put c×a=n ∴[a×bb×cc×a]=(a×b)⋅{(b×c)×n} =(a×b)⋅{(n⋅b)c−(n⋅c)b} =(a×b)⋅[{(c×a)⋅b}c−{(c×a)⋅c}b] =(a×b)⋅{[cab]c−[cac]b} =[(a×b)⋅c][cab]−0 =[abc][abc] =[abc]2=(4)2=16
Alternate Method
By properties of Scalar triple product, [a×bb×cc×a]=[abc]2 =(4)2=16