If | … | =α(a+b() b+c() c+a) ≠ 0 then α is equal to

Let $\Delta=\left|\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}\right|$Δ=​−2ab+ac+a​a+b−2bb+c​a+cb+c−2c​​ Applying $C_1+C_3$C1​+C3​ and $C_2+C_3$C2​+C3​ $$\Delta=\left|\begin{array}{ccc} -a+c & 2 a+b+c & a+c \\ 2 b+a+c & -b+c & b+c \\ a-c & b-c & -2 c \end{array}\right|$$Δ=​−a+c2b+a+ca−c​2a+b+c−b+cb−c​a+cb+c−2c​​ Now, applying $R_1+R_3$R1​+R3​ and $R_2+R_3$R2​+R3​ $$\Delta=\left|\begin{array}{ccc} 0 & 2(a+b & ) a-c \\ 2(a+b & 0 & b-c \\ a-c & b-c & -2 c \end{array}\right|$$Δ=​02(a+ba−c​2(a+b0b−c​)a−cb−c−2c​​ On expanding, we get $$\begin{aligned} \Delta= & -2(a+b)\{-2 c[2(a+b)] \\ & -(a-c)(b-c)\} \\ & +(a-c)[2(a+b)(b-c)] \\ \Delta= & 8 \mathrm{c}(a+b)(a+b) \\ & +4(a+b)(a-c)(b-c) \end{aligned}$$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)​ $$\begin{aligned} & =4(a+b)\left[2 a c+2 b c+a b-b c-a c+c^2\right] \\ & =4(a+b)\left[a c+b c+a b+c^2\right] \\ & =4(a+b)[c(a+c)+b(a+c)] \\ & =4(a+b)(b+c)(c+a) \\ & =\alpha(a+b)(b+c)(c+a) \end{aligned}$$​=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, $\alpha=4$α=4