MHTCET 16 May 2024 Shift 2 Paper

Concept:The scalar triple product $[\bar{a}\,\bar{b}\,\bar{c}]$ expands linearly, and cyclic permutations of its vectors leave the value unchanged.Explanation:On the RHS, $\bar{a}\cdot\hat{i}=a_{1}$, $\bar{a}\cdot\hat{j}=a_{2}$, $\bar{a}\cdot\hat{k}=a_{3}$, and similarly for $\bar{b}$ and $\bar{c}$.So the given determinant equals $\begin{vmatrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{vmatrix} = [\bar{a}\,\bar{b}\,\bar{c}]$.Now expand the LHS as a scalar triple product:$[3\bar{a}+\bar{b}\quad 3\bar{b}+\bar{c}\quad 3\bar{c}+\bar{a}] = (3\bar{a}+\bar{b})\cdot\big((3\bar{b}+\bar{c})\times(3\bar{c}+\bar{a})\big)$.First compute the cross product:$(3\bar{b}+\bar{c})\times(3\bar{c}+\bar{a}) = 9(\bar{b}\times\bar{c}) + 3(\bar{b}\times\bar{a}) + 3(\bar{c}\times\bar{c}) + (\bar{c}\times\bar{a})$.Since $\bar{c}\times\bar{c}=\bar{0}$, this simplifies to $9(\bar{b}\times\bar{c}) + 3(\bar{b}\times\bar{a}) + (\bar{c}\times\bar{a})$.Taking the dot product with $3\bar{a}+\bar{b}$ gives:$27[\bar{a}\,\bar{b}\,\bar{c}] + [\bar{b}\,\bar{c}\,\bar{a}]$.Using the cyclic property $[\bar{b}\,\bar{c}\,\bar{a}] = [\bar{a}\,\bar{b}\,\bar{c}]$, the LHS becomes:$27[\bar{a}\,\bar{b}\,\bar{c}] + [\bar{a}\,\bar{b}\,\bar{c}] = 28[\bar{a}\,\bar{b}\,\bar{c}]$.Comparing with $\lambda[\bar{a}\,\bar{b}\,\bar{c}]$, we get $\lambda = 28$.Answer:$\lambda = 28$, so the correct option is B.