If a→=5i^−j^+k^,b→=2i^−3j^−k^,c→=−3i^+j^+k^ and d→=2j^+k^, then  the value of d→⋅(a→×{b→×(c→×d→)}) equals

Given vectors are a→=5i^−j^+k^,b→=2i^−3j^−k^,c→=−3i^+j^+k^ and d→=2j^+k^ We have a→×(b→×c→)=(a→⋅c→)b→−(a→⋅b→)c→ Here b→⋅d→=−7 and b→⋅c→=−10 Now, b→×(c→×d→)=(b→⋅d→)c→−(b→⋅c→)d→=−7c→+10d→a→×(b→×(c→×d→))=a→×(10d→−7c→)=10a→×d→−7a→×c→ Now, d→⋅(a→×(b→×(c→×d→)))=d→⋅(10a→×d→−7a→×c→)=10[d→a→d→]−7[d→ a→c→]=10(0)−70215−11−311=−7[0−2(5+3)+1(5−3)]=98Therefore, the correct answer is (1).