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(B) Let $\Delta = \left| \begin{matrix} a - b - c & 2a & 2a \ 2b & b - c - a & 2b \ 2c & 2c & c - a - b \end{matrix} ight|$.Applying the row opera

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$-2(a + b + c)$

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(B) Let $\Delta = \left| \begin{matrix} a - b - c & 2a & 2a \ 2b & b - c - a & 2b \ 2c & 2c & c - a - b \end{matrix} ight|$.Applying the row operation $R_1 	o R_1 + R_2 + R_3$:$\Delta = \left| \begin{matrix} a + b + c & a + b + c & a + b + c \ 2b & b - c - a & 2b \ 2c & 2c & c - a - b \end{matrix} ight|$.Taking $(a + b + c)$ common from $R_1$:$\Delta = (a + b + c) \left| \begin{matrix} 1 & 1 & 1 \ 2b & b - c - a & 2b \ 2c & 2c & c - a - b \end{matrix} ight|$.Applying column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\Delta = (a + b + c) \left| \begin{matrix} 1 & 0 & 0 \ 2b & -(a + b + c) & 0 \ 2c & 0 & -(a + b + c) \end{matrix} ight|$.Expanding along $R_1$:$\Delta = (a + b + c) [1 \cdot (-(a + b + c)) \cdot (-(a + b + c)) - 0] = (a + b + c)^3$.Given $\Delta = (a + b + c)(x + a + b + c)^2$,we have:$(a + b + c)^3 = (a + b + c)(x + a + b + c)^2$.Since $a + b + c 
e 0$,we divide by $(a + b + c)$:$(a + b + c)^2 = (x + a + b + c)^2$.Taking the square root on both sides:$x + a + b + c = \pm(a + b + c)$.If $x + a + b + c = a + b + c$,then $x = 0$ (rejected as $x 
e 0$).If $x + a + b + c = -(a + b + c)$,then $x = -2(a + b + c)$.

If $\left| \begin{matrix} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{matrix} \right| = (a + b + c)(x + a + b + c)^2$,$x \ne 0$ and $a + b + c \ne 0$,then $x$ is equal to

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