a,b,c are there non-coplanar vectors (a−λb).[(b−2c)×(c+2a)]=0
⇒(a−λb).{b×c+2(b×a)−2(c×c)−4(c×a)}=0
⇒(a−λb)−b×c+2(b×a)−4(c×a)}=0[∵a×a=0] ⇒a.(b×c)+2a.(b×a)−4a.(c×a) −λb.(b×c)−2λb.(b×a)+4λb.(c×a)=0 ⇒[abc]+2aba]−4[aca]−λ[bbc] −2λ[bba]+4λ|bca|=0 ∴ We know that in scalar triple product, if two vectors are same in box then its value is 0.