Let Δ=−2ab+ac+aa+b−2bb+ca+cb+c−2c Applying C1+C3 and C2+C3 Δ=−a+c2b+a+ca−c2a+b+c−b+cb−ca+cb+c−2c Now, applying R1+R3 and R2+R3 Δ=02(a+ba−c2(a+b0b−c)a−cb−c−2c On expanding, we get Δ=Δ=−2(a+b){−2c[2(a+b)]−(a−c)(b−c)}+(a−c)[2(a+b)(b−c)]8c(a+b)(a+b)+4(a+b)(a−c)(b−c) =4(a+b)[2ac+2bc+ab−bc−ac+c2]=4(a+b)[ac+bc+ab+c2]=4(a+b)[c(a+c)+b(a+c)]=4(a+b)(b+c)(c+a)=α(a+b)(b+c)(c+a) Hence, α=4