Given, a=(1,2,3)=i+2j+3k b=(2,−1,1)=2i−j+k
and c=(3,2,1)=3i+2j+k
Now, a×(b×c)=αa+βb+γc ⇒(a⋅c)b−(a⋅b)c=αa+βb+γc ⇒{(i+2j+3k)⋅(3i+2j+k)}b −{(i+2j+3k)⋅(2i−j+k)}c =αa+βb+γc ⇒(3+4+3)b−(2−2+3)c=αa+βb+γc ⇒0a+10b−3c=αa+βb+γc
On comparing, we get α=0,β=10 and γ=−3