Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three vectors having magnitudes 1,2,3 respectively satisfy the relation $\left|\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]\right|=6$ . If  $\overrightarrow{d}$ is a unit vector coplanar with $\overrightarrow{b}and\overrightarrow{c}$   such that $\overrightarrow{b}.\overrightarrow{d}=1$ then the value of   ${|(\overrightarrow{a}\times \overrightarrow{c}).\overrightarrow{d}|}^2-{|(\overrightarrow{a}\times \overrightarrow{c})\times \overrightarrow{d}|}^2$

Let the angle between $\overrightarrow{a}and\overrightarrow{b}$ $is\alpha and\overrightarrow{a}\times \overrightarrow{b}and\overrightarrow{c}is\beta$  
$\begin{array}{l}\therefore \left|[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]\right|=6\Rightarrow \sin \alpha \cos \beta =1\\ So\Rightarrow \sin \alpha =1,\cos \beta =1\Rightarrow \alpha ={90}^0,\beta =0^0\end{array}$ 
$\Rightarrow \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$  are mutually perpendicular
$again{\left[\overrightarrow{b}\overrightarrow{c}\overrightarrow{d}\right]}^2=0\Rightarrow \left|\begin{array}{ccc}4& 0& 1\\ 0& 9& \overrightarrow{c}.\overrightarrow{d}\\ 1& \overrightarrow{c}.\overrightarrow{d}& 1\end{array}\right|=0$ 
$\Rightarrow \overrightarrow{c}.\overrightarrow{d}=\pm \frac{3\sqrt{2}}{2}$  we have   $\overrightarrow{a}.\overrightarrow{d}=0$
${|\overrightarrow{a}\times \overrightarrow{c}.\overrightarrow{d}|}^2=\left|\begin{array}{ccc}1& 0& 0\\ 0& 9& \frac{3\sqrt{3}}{2}\\ 0& \frac{3\sqrt{3}}{2}& 1\end{array}\right|=9-\frac{27}{4}=\frac{9}{4}$ 
 ${|(\overrightarrow{a}\times \overrightarrow{c}).\times \overrightarrow{d}|}^2$
 $={|(\overrightarrow{a}.\overrightarrow{d})\overrightarrow{c}-(\overrightarrow{c}.\overrightarrow{d})\overrightarrow{a}|}^2={\left|\frac{3\sqrt{3}}{2}\overrightarrow{a}\right|}^2=\frac{27}{4}$
$\begin{array}{l}\therefore {|\overrightarrow{a}\times \overrightarrow{c}.\overrightarrow{d}|}^2-{|(\overrightarrow{a}\times \overrightarrow{c})\times \overrightarrow{d}|}^2\\ =\frac{9}{4}-\frac{27}{4}=-\frac{18}{4}=-\frac{9}{2}\\ \end{array}$