If a→=2i^+j^+k^,b→=i^+2j^+2k^,c→=i^+j^+2k^ and (1+α)i^+β(1+α)j^+γ(1+α)(1+β)k^ =a→×(b→×c→) then α,β and γ are
−2,−4,−23
2,−4,23
−2,4,23
2,4,−23
a→×(b→×c→)=(a→⋅c→)b→−(a→⋅b→)c→=5(i^+2j^+2k^)−6(i^+j^+2k^)⇒(1+α)i^+β(1+α)j^+γ(1+α)(1+β)k^=−i^+4j^−2k^⇒1+α=−1,β=−4 and γ(−1)(−3)=−2⇒ γ=−23