If x = cy + bz, y = az + cx, z = bx + ay where x, y, z are not all zeros, then the value of $a^{2} + b^{2} + c^{2} + 2 a b c$ is

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Solution

The correct option is B
1

Given system x - cy - bz = 0, cx - y + az = 0, bx + ay - z = 0
Since the system of equations has non trivial solution,
$∣ ∣ ∣ ∣ \begin{matrix} 1 & - c & - b c & - 1 & a b & a & - 1 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow 1 - a b c - a b c - b^{2} - a^{2} - c^{2} = 0 \Rightarrow a^{2} + b^{2} + c^{2} + 2 a b c = 1$

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