Let three vectors → a , → b and → c be such that → a × → b = → c ,...

The first option is $\overset{\to}{a} \times \left(\left(\overset{\to}{b} + \overset{\to}{c}\right) \times \left(\overset{\to}{b} - \overset{\to}{c}\right)\right)$

$= \overset{\to}{a} \times \left(\left(\overset{\to}{b} \times \overset{\to}{b} - \overset{\to}{b} \times \overset{\to}{c} + \overset{\to}{c} \times \overset{\to}{b} - \overset{\to}{c} \times \overset{\to}{c}\right)\right)$

But, we know that $\overset{\to}{b} \times \overset{\to}{b} = 0 = \overset{\to}{c} \times \overset{\to}{c}$

$= \overset{\to}{a} \times \left(\overset{\to}{c} \times \overset{\to}{b} - \overset{\to}{b} \times \overset{\to}{c}\right)$

Given, $\overset{\to}{b} \times \overset{\to}{c} = \overset{\to}{a}, ⇒ \overset{\to}{c} \times \overset{\to}{b} = - \overset{\to}{a}$

$= \overset{\to}{a} \times \left(- \overset{\to}{a} - \overset{\to}{b} \times \overset{\to}{c}\right)$

$= \left(- \overset{\to}{a} \times \overset{\to}{a}\right) - \left(\overset{\to}{a} \times \left(\overset{\to}{b} \times \overset{\to}{c}\right)\right)$

$= 0 - \left(\overset{\to}{a} \times \left(\overset{\to}{b} \times \overset{\to}{c}\right)\right) = - \left(\overset{\to}{a} \times \left(\overset{\to}{b} \times \overset{\to}{c}\right)\right)$

$= - ( \overset{\to}{a} \times \overset{\to}{a} ) = \overset{\to}{0}$

$∵ \overset{\to}{b} \times \overset{\to}{c} = \overset{\to}{a}$

Hence, the option \left(A\right)$is correct.</p><p>For the second option, we have </p><p>Projection of$\overset{\to}{a}$on$\overset{\to}{b} \times \overset{\to}{c}$is</p><p>$= \dfrac{\overset{\to}{a} \cdot ( \overset{\to}{b} \times \overset{\to}{c} )}{| \overset{\to}{b} \times \overset{\to}{c} |} = \dfrac{\overset{\to}{a} \cdot \overset{\to}{a}}{| \overset{\to}{a} |}$</p><p>Using,$\overset{\to}{a} \cdot \overset{\to}{a} = {\left|\overset{\to}{a}\right|}^{2},$we get</p><p>$= \dfrac{{\left|\overset{\to}{a}\right|}^{2}}{\left|\overset{\to}{a}\right|} = \left|\overset{\to}{a}\right| = 2 .$</p><p>Hence, the option \left(B\right)$ is correct.

For the third option, we have \left[$\begin{cases}\overset{\to}{a} & \overset{\to}{b} & \overset{\to}{c}\end{cases}$\right] + \left[$\begin{cases}\overset{\to}{c} & \overset{\to}{a} & \overset{\to}{b}\end{cases}$\right]</p><p>We know that \left[\begin{cases}\overset{\to}{a} & \overset{\to}{b} & \overset{\to}{c}\end{cases}\right] = \left[\begin{cases}\overset{\to}{c} & \overset{\to}{a} & \overset{\to}{b}\end{cases}\right], and \left[\begin{cases}\overset{\to}{a} & \overset{\to}{b} & \overset{\to}{c}\end{cases}\right] = \overset{\to}{a} \cdot ( \overset{\to}{b} \times \overset{\to}{c} ) ,

⇒ \left[\begin{cases}\overset{\to}{a} & \overset{\to}{b} & \overset{\to}{c}\end{cases}\right] + \left[\begin{cases}\overset{\to}{c} & \overset{\to}{a} & \overset{\to}{b}\end{cases}\right] = 2\left[\begin{cases}\overset{\to}{a} & \overset{\to}{b} & \overset{\to}{c}\end{cases}\right] = 2\overset{\to}{a} \cdot ( \overset{\to}{b} \times \overset{\to}{c} )

$= 2\left(\overset{\to}{a} \cdot \overset{\to}{a}\right) = 2 | \overset{\to}{a}{|}^{2} = 2 \times {2}^{2} = 8 .$

Hence, the option \left(C\right)$is correct.</p><p>For the fourth option, we have</p><p>$\overset{\to}{a} \times \overset{\to}{b} = \overset{\to}{c}$and$\overset{\to}{b} \times \overset{\to}{c} = \overset{\to}{a}$</p><p>$⇒ \overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c}$are mutually perpendicular vectors.</p><p>$∴ | \overset{\to}{a} \times \overset{\to}{b} | = | \overset{\to}{c} | ⇒ \left|\overset{\to}{a}\right|\left|\overset{\to}{b}\right| = \left|\overset{\to}{c}\right|$</p><p>$⇒ \left|\overset{\to}{b}\right| = \dfrac{\left|\overset{\to}{c}\right|}{2} . . . \left(i\right)$</p><p>Also,$| \overset{\to}{b} \times \overset{\to}{c} | = | \overset{\to}{a} |$</p><p>$⇒ | \overset{\to}{b} | | \overset{\to}{c} | = 2$</p><p>On putting the value from equation \left(i\right),$ we get

$⇒ \left(\dfrac{| \overset{\to}{c} |}{2}\right) | \overset{\to}{c} | = 2$

$⇒ | \overset{\to}{c}{|}^{2} = 4$

$⇒ | \overset{\to}{c} | = 2$ and $| \overset{\to}{b} | = 1$

$| 3\overset{\to}{a} + \overset{\to}{b} - 2\overset{\to}{c}{|}^{2} = ( 3\overset{\to}{a} + \overset{\to}{b} - 2\overset{\to}{c} ) \cdot ( 3\overset{\to}{a} + \overset{\to}{b} - 2\overset{\to}{c} )$

The vectors are mutually perpendicular to each other, hence

$| 3\overset{\to}{a} + \overset{\to}{b} - 2\overset{\to}{c}{|}^{2} = 9 | \overset{\to}{a}{|}^{2} + | \overset{\to}{b}{|}^{2} + 4 | \overset{\to}{c}{|}^{2}$ \left[∵ \overset{\to}{a} \cdot \overset{\to}{b} = \overset{\to}{b} \cdot \overset{\to}{c} = \overset{\to}{c} \cdot \overset{\to}{a} = 0\right]$</p><p>$= ( 9 \times 4 ) + 1 + ( 4 \times 4 )$</p><p>$= 36 + 1 + 16 = 53 .$</p><p>Hence, the option \left(D\right)$ is not correct.