|ab+c1ba+c1ca+b1| C1+C2+C3 =|a+b+ca+c1a+b+ca+c1a+b+ca+b1|=(a+b+c)|1b+c11a+c11a+b1|=0

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$| \begin{matrix} a & b + c & 1 \\ b & a + c & 1 \\ c & a + b & 1 \end{matrix} | =$

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$a + b + c$

${(a + b + c)}^{3}$

${(a + b + c)}^{2}$

answer is C.

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Detailed Solution

$| \begin{matrix} a & b + c & 1 \\ b & a + c & 1 \\ c & a + b & 1 \end{matrix} | C_{1} + C_{2} + C_{3} = | \begin{matrix} a + b + c & a + c & 1 \\ a + b + c & a + c & 1 \\ a + b + c & a + b & 1 \end{matrix} | = (a + b + c) | \begin{matrix} 1 & b + c & 1 \\ 1 & a + c & 1 \\ 1 & a + b & 1 \end{matrix} | = 0$