If → a , → b , → c are non-coplaner vectors such that → b × → c = →...

Given,&

$\overset{→}{a} \times \overset{→}{b} = \overset{→}{c}$

$\overset{→}{b} \times \overset{→}{c} = \overset{→}{a}$

$\overset{→}{c} \times \overset{→}{a} = \overset{→}{b}$

$( \overset{→}{b} \times \overset{→}{c} ) \cdot \overset{→}{a} = \overset{→}{a} \cdot \overset{→}{a}$

$( \overset{→}{c} \times \overset{→}{a} ) \cdot \overset{→}{b} = \overset{→}{b} \cdot \overset{→}{b}$

$\overset{→}{a}2 - \overset{→}{b}2 = ( \overset{→}{b} \times \overset{→}{c} ) \cdot \overset{→}{a} - ( \overset{→}{c} \times \overset{→}{a} ) \cdot \overset{→}{b}$

$= [$\begin{cases}\overset{→}{c} & \overset{→}{a} & \overset{→}{b}\end{cases}$] - [$\begin{cases}\overset{→}{b} & \overset{→}{c} & \overset{→}{a}\end{cases}$] = 0$

$∴ | \overset{→}{a}{|}^{2} - | \overset{→}{b}{|}^{2} = 0$

$| \overset{→}{a} | - | \overset{→}{b} | = 0$

$⇒ | \overset{→}{a} | = | \overset{→}{b} |$

We know that,~

\left[\overset{→}{a} \times \overset{→}{b}\overset{→}{b} \times \overset{→}{c}\overset{→}{c} \times \overset{→}{a}\right] = [ \overset{→}{a}\overset{→}{b}\overset{→}{c}{]}^{2}</p><p>\left[$\begin{array}{ll}\overset{→}{a} \overset{→}{b} \overset{→}{c}\end{array}$$\right] = \left[\begin{cases}\overset{→}{a} & \overset{→}{b} & \overset{→}{c}\end{cases}\right]2$</p><p>$⇒ \left[$\begin{cases}\overset{→}{a} & \overset{→}{b} & \overset{→}{c}\end{cases}$\right]$= 1$

$| \overset{→}{a} | | \overset{→}{b} | | \overset{→}{c} | = 1$

$∴ | \overset{→}{a} | = | \overset{→}{b} | = | \overset{→}{c} | \ne 2$