Match List-I with List - II and select the correct answer using the code given below the lists List-I…

(A)   Given, $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=2 ...\left(1\right)$

 The volume of the parallelepiped determined by vectors

$2\left(\overrightarrow{a}\times \overrightarrow{b}\right),3\left(\overrightarrow{b}\times \overrightarrow{c}\right),\left(\overrightarrow{c}\times \overrightarrow{a}\right)$  is the scalar tripple product of given three vectors is
$\Rightarrow V=\left[2\left(\overrightarrow{a}\times \overrightarrow{b}\right)3\left(\overrightarrow{b}\times \overrightarrow{c}\right)\left(\overrightarrow{c}\times \overrightarrow{a}\right)\right] ...\left(2\right)$

$\Rightarrow V=6\left[\left( \overrightarrow{a}\times \overrightarrow{b}\right)\left(\overrightarrow{b}\times \overrightarrow{c}\right)\left(\overrightarrow{c}\times \overrightarrow{a}\right)\right]$

 Using this result ,  $\left[\left(\overrightarrow{a }\times \overrightarrow{b}\right) \left(\overrightarrow{b }\times \overrightarrow{c}\right) \left(\overrightarrow{c }\times \overrightarrow{a}\right)\right]={\left[ \overrightarrow{a} \overrightarrow{b} \overrightarrow{c}\right]}^2 ...\left(3\right)$

$\Rightarrow V=6{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}^2$

Again putting the value of $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=2$

We get the volume of the parallelepiped, $V=6\times 4=24$.
$\text{A→R}$
(B) Given, $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=5 ...\left(1\right)$

$V=\left[3\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{b}+\overrightarrow{c}\right)2\left(\overrightarrow{c}+\overrightarrow{a}\right)\right]$

$\Rightarrow V=6\left[\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{b}+\overrightarrow{c}\right)\left(\overrightarrow{c}+\overrightarrow{a}\right)\right] ...\left(2\right)$

Using results, $\left[\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{b}+\overrightarrow{c}\right)\left(\overrightarrow{c}+\overrightarrow{a}\right)\right]=2\left[\left(\overrightarrow{a}\right)\left(\overrightarrow{b}\right)\left(\overrightarrow{c}\right)\right] ...\left(3\right)$
$\Rightarrow V=6\times 2\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$

$\Rightarrow V=12 \left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$
Putting the value $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$ from equation $\left(1\right)$, we get the volume of the parallelepiped ,

 $\Rightarrow V=12\times 5=60$.
$\text{B⟶S}$
(C))  Given,Area of triangle whose adjacents sides are $\overrightarrow{a} and \overrightarrow{b}$,

 $\Rightarrow ∆=\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{b}\right|=20 ...\left(1\right)$
${\Delta }_1=\frac{1}{2}\left|\left(2\overrightarrow{a}+3\overrightarrow{b}\right)\times \left(\overrightarrow{a}-\overrightarrow{b}\right)\right|$
$\Rightarrow {∆}_1=\frac{1}{2}\left|-2\left(\overrightarrow{a}\times \overrightarrow{b}\right)-3\left(\overrightarrow{a}\times \overrightarrow{b}\right)\right|$
$\Rightarrow {∆}_1=5 \left(\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{b}\right|\right)$
Using  the results from equation $\left(1\right)$

$\Rightarrow {∆}_1=5\times 20=100$
$\text{C⟶P}$
(D) Given, The area of the parallelogram with adjacent sides determined by vectors $\left(\overrightarrow{a}\right)$ and $\left(\overrightarrow{b}\right)$

$\left|\overrightarrow{a}\times \overrightarrow{b}\right|=30 ...\left(1\right)$
 The area of the parallelogram with adjacent sides determined by vectors  $\left(\overrightarrow{a}+\overrightarrow{b}\right)$ and $\left(\overrightarrow{a}\right)$                                     

$\Rightarrow \mathrm{Area} A=\left|\left(\overrightarrow{a}+\overrightarrow{b}\right)\times \left(\overrightarrow{a}\right)\right|=\left|\overrightarrow{b}\times \overrightarrow{a}\right|$, Because $\left(\overrightarrow{a}\times \overrightarrow{a}\right)=0$

$\Rightarrow \mathrm{Area} A=\left|\overrightarrow{b}\times \overrightarrow{a}\right|$

Using results from $\left(1\right)$,we get

$\Rightarrow \mathrm{Area} A=\left|\overrightarrow{b}\times \overrightarrow{a}\right|=30$
$\text{D⟶Q}$