$∣ ∣ ∣ ∣ \begin{matrix} 2 c & 2 c & c - a - b a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b \end{matrix} ∣ ∣ ∣ ∣$ =

(a + b + c)^{2}

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(a + b + c)^{3}

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(a + b + c)

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(a + b + c)^{4}

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Solution

The correct option is A $(a + b + c)^{3}$
$∣ ∣ ∣ ∣ \begin{matrix} 2 c & 2 c & c - a - b a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b \end{matrix} ∣ ∣ ∣ ∣$
$C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3}$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & (a + b + c) & c - a - b - (a + b + c) & 0 & 2 a (a + b + c) & - (a + b + c) & 2 b \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= {(a + b + c)}^{2} ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & c - a - b - 1 & 0 & 2 a 1 & - 1 & 2 b \end{matrix} ∣ ∣ ∣ ∣$
$R_{3} \to R_{3} + R_{2}$
$= {(a + b + c)}^{2} ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & c - a - b - 1 & 0 & 2 a 0 & - 1 & 2 a + 2 b \end{matrix} ∣ ∣ ∣ ∣$
$= {(a + b + c)}^{2} [1 (2 a + 2 b + c - a - b)]$
$= {(a + b + c)}^{3}$

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