If $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$=(a+b+c)(x+a+b+c)^{2}, x \neq 0$ and $a+b+c \neq 0$, then $x$ is equal to :-
Correct Option: , 4
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$
$=\left|\begin{array}{ccc}a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$=(a+b+c)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 b & -(a+b+c) & 0 \\ 2 c & 2 c & c-a-b\end{array}\right|$
$=(a+b+c)(a+b+c)^{2}$
$\Rightarrow x=-2(a+b+c)$
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