If $a x_{1}^{2} + b y_{1}^{2} + c z_{1}^{2} = a x_{2}^{2} + b y_{2}^{2} + c z_{2}^{2} = a x_{3}^{2} + b y_{3}^{2} + c z_{3}^{2} = d$ and $a x_{2} x_{3} + b y_{2} y_{3} + c z_{2} z_{3} = a x_{3} x_{1} + b y_{3} y_{1} + c z_{3} z_{1} = a x_{1} x_{2} + b y_{1} y_{2} + c z_{1} z_{2} = f$ , then prove that $∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & z_{1} x_{2} & y_{2} & z_{2} x_{3} & y_{3} & z_{3} \end{matrix} ∣ ∣ ∣ ∣ = (d - f) {[\frac{d + 2 f}{a b c}]}^{1 / 2} (a, b, c \neq 0)$

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Solution

Given ,
$a x_{1}^{2} + b y_{1}^{2} + c z_{1}^{2} = a x_{2}^{2} + b y_{2}^{2} + c z_{2}^{2} = a x_{3}^{2} + b y_{3}^{2} + c z_{3}^{2} = d$ .....(1) $a x_{2} x_{3} + b y_{2} y_{3} + c z_{2} z_{3} = a x_{3} x_{1} + b y_{3} y_{1} + c z_{3} z_{1} = a x_{1} x_{2} + b y_{1} y_{2} + c z_{1} z_{2} = f$ ....(2)
Let $Δ = ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & z_{1} x_{2} & y_{2} & z_{2} x_{3} & y_{3} & z_{3} \end{matrix} ∣ ∣ ∣ ∣$
$Δ^{2} = \frac{1}{a b c} ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & z_{1} x_{2} & y_{2} & z_{2} x_{3} & y_{3} & z_{3} \end{matrix} ∣ ∣ ∣ ∣ \times ∣ ∣ ∣ ∣ \begin{matrix} a x_{1} & b y_{1} & c z_{1} a x_{2} & b y_{2} & c z_{2} a x_{3} & b y_{3} & c z_{3} \end{matrix} ∣ ∣ ∣ ∣$
$= \frac{1}{a b c} ∣ ∣ ∣ ∣ \begin{matrix} d & f & f f & d & f f & f & d \end{matrix} ∣ ∣ ∣ ∣$ (by (1) and (2))
$C_{1} \to C_{1} + C_{2} + C_{3}$
$= \frac{1}{a b c} ∣ ∣ ∣ ∣ \begin{matrix} d + 2 f & f & f d + 2 f & d & f d + 2 f & f & d \end{matrix} ∣ ∣ ∣ ∣ = \frac{(d + 2 f)}{a b c} ∣ ∣ ∣ ∣ \begin{matrix} 1 & f & f 1 & d & f 1 & f & d \end{matrix} ∣ ∣ ∣ ∣$
On solving it we get
$Δ^{2} = \frac{(d + 2 f)}{a b c} (d - f)^{2}$
$Δ = (d - f) {[\frac{d + 2 f}{a b c}]}^{\frac{1}{2}}$

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a x_{1}^{2} + b y_{1}^{2} + c z_{1}^{2} = a x_{2}^{2} + b y_{2}^{2} + c z_{2}^{2} = a x_{3}^{2} + b y_{3}^{2} + c z^{3} = d, a x_{2} x_{3} + b y_{2} y_{3} + c z_{2} z_{3} = a x_{3} x_{1} + b y_{3} y_{1} + c z_{3} z_{1} = a x_{1} x_{2} + b y_{1} y_{2} + c z_{1} z_{2} = f

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