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(B) Let $\Delta = \left|\begin{array}{ccc} a+b+2c & a & b \ c & b+c+2a & b \ c & a & c+a+2b \end{array}ight|$.Applying the column operation $C_1 \

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$2(a+b+c)^3$

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(B) Let $\Delta = \left|\begin{array}{ccc} a+b+2c & a & b \ c & b+c+2a & b \ c & a & c+a+2b \end{array}ight|$.Applying the column operation $C_1 ightarrow C_1 + C_2 + C_3$,we get:$\Delta = \left|\begin{array}{ccc} 2(a+b+c) & a & b \ 2(a+b+c) & b+c+2a & b \ 2(a+b+c) & a & c+a+2b \end{array}ight|$.Taking $2(a+b+c)$ common from $C_1$:$\Delta = 2(a+b+c) \left|\begin{array}{ccc} 1 & a & b \ 1 & b+c+2a & b \ 1 & a & c+a+2b \end{array}ight|$.Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$\Delta = 2(a+b+c) \left|\begin{array}{ccc} 1 & a & b \ 0 & a+b+c & 0 \ 0 & 0 & a+b+c \end{array}ight|$.Expanding along $C_1$,we get:$\Delta = 2(a+b+c) \cdot (1) \cdot [(a+b+c)(a+b+c) - 0] = 2(a+b+c)^3$.

$\left|\begin{array}{ccc} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{array}\right| = $

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