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If ∣(a−x)2(a−y)2(a−z)2(b−x)2(b−y)2(b−z)2(c−x)2(c−y)2(c−z)2∣=λ(a−b)(b−c)(c−a)⋅(x−y)(y−z)(z−x)\begin{vmatrix} (a-x)^2 & (a-y)^2 & (a-z)^2 \\ (b-x)^2 & (b-y)^2 & (b-z)^2 \\ (c-x)^2 & (c-y)^2 & (c-z)^2 \end{vmatrix} = \lambda(a-b)(b-c)(c-a) \cdot (x-y)(y-z)(z-x)

then the value of λ\lambda is:

Solution

✅ Correct Option: 2

Test with a=1,b=0,c=2,x=0,y=1,z=2a=1, b=0, c=2, x=0, y=1, z=2: determinant evaluates to −8-8. Right side without λ\lambda: (1)(−2)(1)(−1)(−1)(2)=−4(1)(−2)(1)(−1)(−1)(2) = -4. So λ=−8/−4=2\lambda = -8 / -4 = 2.

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