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

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$9 b^{2} (a + b)$

$9 a^{2} (a + b)$

$9 (a + b)^{3}$

$9 ab (a + b)$

answer is A.

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Detailed Solution

$|\begin{array}{ccc} a & a + b & a + 2 b \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array}|$
$Now apply C_{1} \to C_{1} + C_{2} + C_{3}$
$= |\begin{array}{ccc} 3 a + 3 b & a + b & a + 2 b \\ 3 a + 3 b & a & a + b \\ 3 a + 3 b & a + 2 b & a \end{array}|$
$= 3 (a + b) |\begin{array}{ccc} 1 & a + b & a + 2 b \\ 1 & a & a + b \\ 1 & a + 2 b & a \end{array}|$
$R_{1} \to R_{1} - R_{3}, R_{2} \to R_{2} - R_{3}$
$= 3 (a + b) |\begin{array}{ccc} 0 & - b & 2 b \\ 0 & - 2 b & a + b \\ 1 & a + 2 b & a \end{array}|$
$\begin{array}{l} \begin{array}{ll} = & 3 (a + b) |\begin{array}{cc} - b & 2 b \\ - 2 b & b \end{array}| \begin{array}{c} \end{array} \\ = & 3 b^{2} (a + b) |\begin{array}{cc} - 1 & 2 \\ 2 & 1 \end{array}| \end{array} \\ = 3 b^{2} (a + b) (- 1 + 4) = 9 b^{2} (a + b) \end{array}$