If \begin{vmatrix} a & -b & -c \ 2a & 2a & 2b \ b & -c & -a \end{vmatrix} = (a+b+c)(x+a+b+c)^2, x \neq 0 and a+b+c \neq 0, then x is equal to:

\begin{vmatrix} a & -b & -c \ 2a & 2a & 2b \ b & -c & -a \end{vmatrix} \begin{aligned} R_1 &\rightarrow R_1 + R_2 + R_3 \ &= \begin{vmatrix} a+b+c & a+b+c & a+b+c \ 2a & 2a & 2b \ b & -c & -a \end{vmatrix} \end{aligned} \text{Take } a+b+c \text{ common from first row} \ C_2 \rightarrow C_2 - C_1, C_3 \rightarrow C_3 - C_2 \ = (a+b+c) \begin{vmatrix} 1 & 0 & 0 \ 2a & 0 & 2b-(2a) \ b & -c-b & -a-(-c) \end{vmatrix} \ = (a+b+c)(a+b+c)^2 \Rightarrow x = -(a+b+c)