Let the vectors ${\vec{u}}_{1} = \hat{i} + \hat{j} + a \hat{k}, {\vec{u}}_{2} = \hat{i} + b \hat{j} + \hat{k} and {\vec{u}}_{3} = c \hat{i} + \hat{j} + \hat{k}$ . If the vectors ${\vec{v}}_{1} = (a + b) \hat{i} + c \hat{j} + c \hat{k}, {\vec{v}}_{2} = a \hat{i} + (b + c) \hat{j} + a \hat{k} and {\vec{v}}_{3} = b \hat{i} + b \hat{j} + (c + a) \hat{k}$ are also coplanar, then
6(a + b + c) is equal to

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Detailed Solution

$|\begin{array}{lll} a 1 1 \\ 1 b 1 \\ 1 1 c \end{array}| = 0 \Rightarrow abc + 2 - a - b - c = 0 a + b + c = abc + 2 . . . . .(1)$
$|\begin{array}{ccc} a + b & c & c \\ a & b + c & a \\ b & b & a + c \end{array}| = 0,$
$\begin{array}{l} (R_{3} \to R_{3} - (R_{1} + R_{2})) \Rightarrow |\begin{array}{ccc} a + b & c & c \\ a & b + c & a \\ - 2 a & - 2 c & 0 \end{array}| = 0 \\ \Rightarrow a b c = 0 \\ ∴ a + b + c = 2 \\ ∴ 6 (a + b + c) = 12 \end{array}$