If $x^{2} + y^{2} + z^{2} \neq 0,$ and given $x = c y +$ bz, $y = a z +$ cx, and $z = b x +$ ay, then the value of $a^{2} + b^{2} + c^{2} + 2 a b c =$

2

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a + b + c

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1

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a b + b c +

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Solution

The correct option is C $1$
$x - c y - b z = 0$
$y - a z - c x = 0$
$z - b x - c y = 0$
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & - c & - b - c & 1 & - a - b & - a & 1 \end{matrix} ∣ ∣ ∣ ∣ = (1 - a^{2}) + c (- c - a b) - b (a c + b)$
$Δ = 1 - a^{2} - c^{2} - a b c - a b c - b^{2}$
Now, $Δ = 0$ ( $∵$ infinite solution of this equation)
$a^{2} + b^{2} + c^{2} + 2 a b c = 1$

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