If $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$…
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
abc
$-(a+b+c)$
$2(a+b+c)$
$-2(a+b+c)$
Solution
$\Delta=\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$R_{1} \rightarrow R_{1}+R_{2}+R$
$\Delta=\left|\begin{array}{ccc}a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$\begin{array}{l}
C_{1} \rightarrow C_{1}-C_{y}, C_{2} \rightarrow C_{2}-C_{3} \\
\Delta=(a+b+c)\left|\begin{array}{ccc}
0 & 0 & 1 \\
0 & -b-c-a & 2 b \\
c+a+b & c+a+b & c-a-b
\end{array}\right|
\end{array}$
$=(a+b+c)(a+b+c)^{2}$
Hence, $x=-2(a+b+c)$