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

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

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

$2 {(a - b - c)}^{3}$

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

answer is D.

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Detailed Solution

The given determinant is $|\begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix}| =$

Use elementary row operations

$R_{1} \to R_{1} + R_{2} + R_{3}$

$|\begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix}| ~ |\begin{matrix} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix}|$

Take $a + b + c$ Common from the first row and then use column operations $C_{1} \to C_{1} - C_{2}$ and $C_{2} \to C_{2} - C_{3}$

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