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

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

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

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

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

Given, $\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}, \Rightarrow \overrightarrow{c}\times \overrightarrow{b}=-\overrightarrow{a}$

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

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

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

$=-(\overrightarrow{a}\times \overrightarrow{a})=\overrightarrow{0}$

$\because \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}$

Hence, the option $\left(A\right)$ is correct.

For the second option, we have  

Projection of $\overrightarrow{a}$ on $\overrightarrow{b}\times \overrightarrow{c}$ is

$=\frac{\overrightarrow{a}·(\overrightarrow{b}\times \overrightarrow{c})}{|\overrightarrow{b}\times \overrightarrow{c}|}=\frac{\overrightarrow{a}·\overrightarrow{a}}{\left|\overrightarrow{a}\right|}$

Using, $\overrightarrow{a}·\overrightarrow{a}={\left|\overrightarrow{a}\right|}^2,$ we get

$=\frac{{\left|\overrightarrow{a}\right|}^2}{\left|\overrightarrow{a}\right|}=\left|\overrightarrow{a}\right|=2.$

Hence, the option $\left(B\right)$ is correct.

For the third option, we have $\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]+\left[\begin{array}{ccc}\overrightarrow{c}& \overrightarrow{a}& \overrightarrow{b}\end{array}\right]$

We know that $\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]=\left[\begin{array}{ccc}\overrightarrow{c}& \overrightarrow{a}& \overrightarrow{b}\end{array}\right],$ and $\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]=\overrightarrow{a}·(\overrightarrow{b}\times \overrightarrow{c}),$

$\Rightarrow \left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]+\left[\begin{array}{ccc}\overrightarrow{c}& \overrightarrow{a}& \overrightarrow{b}\end{array}\right]=2\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]=2\overrightarrow{a}·(\overrightarrow{b}\times \overrightarrow{c})$

$=2\left(\overrightarrow{a}·\overrightarrow{a}\right)=2|\overrightarrow{a}{|}^2=2\times 2^2=8.$

Hence, the option $\left(C\right)$ is correct.

For the fourth option, we have

$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}$ and $\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}$

$\Rightarrow \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are mutually perpendicular vectors.

$\therefore |\overrightarrow{a}\times \overrightarrow{b}|=\left|\overrightarrow{c}\right|\Rightarrow \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|$

$\Rightarrow \left|\overrightarrow{b}\right|=\frac{\left|\overrightarrow{c}\right|}{2} ...\left(i\right)$

Also, $|\overrightarrow{b}\times \overrightarrow{c}|=\left|\overrightarrow{a}\right|$

$\Rightarrow \left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|=2$

On putting the value from equation $\left(i\right),$ we get

$\Rightarrow \left(\frac{\left|\overrightarrow{c}\right|}{2}\right)\left|\overrightarrow{c}\right|=2$

$\Rightarrow |\overrightarrow{c}{|}^2=4$

$\Rightarrow \left|\overrightarrow{c}\right|=2$ and  $\left|\overrightarrow{b}\right|=1$

$|3\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c}{|}^2=(3\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c})·(3\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c})$

The vectors are mutually perpendicular to each other, hence 

$|3\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c}{|}^2=9|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+4|\overrightarrow{c}{|}^2$  $\left[\because \overrightarrow{a}·\overrightarrow{b}=\overrightarrow{b}·\overrightarrow{c}=\overrightarrow{c}·\overrightarrow{a}=0\right]$

$=(9\times 4)+1+(4\times 4)$

$=36+1+16=53.$

Hence, the option $\left(D\right)$ is not correct.