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Matrices & Determinants: Determinant Vmatrix Vmatrix Equal

JEE Maths question with a full step-by-step solution.

Question
The determinant ∣0(a−b)2(a−c)2(b−a)20(b−c)2(c−a)2(c−b)20∣\begin{vmatrix} 0 & (a-b)^2 & (a-c)^2 \\ (b-a)^2 & 0 & (b-c)^2 \\ (c-a)^2 & (c-b)^2 & 0 \end{vmatrix} is equal to:
A(a−b)2(b−c)2(c−a)2(a-b)^2(b-c)^2(c-a)^2
B00
C2(a−b)2(b−c)2(c−a)22(a-b)^2(b-c)^2(c-a)^2correct
DNone of these
Solution
Step 1: Expand along row 1
Δ=−(a−b)2∣(b−a)2(b−c)2(c−a)20∣⏟M12+(a−c)2∣(b−a)20(c−a)2(c−b)2∣⏟M13\Delta = -(a-b)^2\underbrace{\begin{vmatrix}(b-a)^2 & (b-c)^2\\(c-a)^2 & 0\end{vmatrix}}_{M_{12}} + (a-c)^2\underbrace{\begin{vmatrix}(b-a)^2 & 0\\(c-a)^2 & (c-b)^2\end{vmatrix}}_{M_{13}}
 The cofactor signs are (−1)1+2=−1(-1)^{1+2} = -1 for the (1,2)(1,2) entry and (−1)1+3=+1(-1)^{1+3} = +1 for the (1,3)(1,3) entry:
M12=0−(b−c)2(c−a)2=−(b−c)2(c−a)2M_{12} = 0 - (b-c)^2(c-a)^2 = -(b-c)^2(c-a)^2
M13=(b−a)2(c−b)2−0=(a−b)2(b−c)2M_{13} = (b-a)^2(c-b)^2 - 0 = (a-b)^2(b-c)^2
 Step 2: Collect terms
Δ=−(−1)(a−b)2(b−c)2(c−a)2+(1)(a−c)2(a−b)2(b−c)2\Delta = -(-1)(a-b)^2(b-c)^2(c-a)^2 + (1)(a-c)^2(a-b)^2(b-c)^2
 Since (a−c)2=(c−a)2(a-c)^2 = (c-a)^2:
Δ=(a−b)2(b−c)2(c−a)2+(a−b)2(b−c)2(c−a)2=2(a−b)2(b−c)2(c−a)2\Delta = (a-b)^2(b-c)^2(c-a)^2 + (a-b)^2(b-c)^2(c-a)^2 = 2(a-b)^2(b-c)^2(c-a)^2
 Answer: (3)
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