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If ∣−1aa2−1bb2−1cc2∣2=λ\begin{vmatrix} -1 & a & a^2 \\ -1 & b & b^2 \\ -1 & c & c^2 \end{vmatrix}^2 = \lambda and a−b=1a - b = 1, b−c=2b - c = 2 and c−a=3c - a = 3, then the value of λ\lambda is

Solution

✅ Correct Option: 2

Taking −1-1 common from column 1 and recognising the Vandermonde structure: ∣1aa21bb21cc2∣=(a−b)(b−c)(c−a)\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = (a-b)(b-c)(c-a). So the determinant equals −(a−b)(b−c)(c−a)-(a-b)(b-c)(c-a) and its square equals (a−b)2(b−c)2(c−a)2=12⋅22⋅32=36(a-b)^2(b-c)^2(c-a)^2 = 1^2 \cdot 2^2 \cdot 3^2 = 36. Hence λ=36\lambda = 36.

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