Match List-I with List - II and select the correct answer using the...

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

The volume of the parallelepiped determined by vectors$</p><p>2 \left(\overset{→}{a} \times \overset{→}{b}\right), 3\left(\overset{→}{b} \times \overset{→}{c}\right), \left(\overset{→}{c} \times \overset{→}{a}\right) is$ the scalar tripple product of given three vectors is
⇒ V = \left[2\left(\overset{→}{a} \times \overset{→}{b}\right)3 \left(\overset{→}{b} \times \overset{→}{c}\right) \left(\overset{→}{c} \times \overset{→}{a}\right)\right] . . . \left(2\right)

$⇒ V = 6 \left[\left(\overset{→}{a} \times \overset{→}{b}\right)\left(\overset{→}{b} \times \overset{→}{c}\right)\left(\overset{→}{c} \times \overset{→}{a}\right)\right]$

Using this result , \left[\left(\overset{→}{a} \times \overset{→}{b}\right)\left(\overset{→}{b} \times \overset{→}{c}\right)\left(\overset{→}{c} \times \overset{→}{a}\right)\right] = \left[\overset{→}{a}\overset{→}{b}\overset{→}{c}\right]2 . . . \left(3\right)$</p><p>$⇒ V = 6 $$\left[\overset{→}{a}\overset{→}{b}\overset{→}{c}\right]$$2

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

We get the volume of the parallelepiped, $V = 6 \times4 = 24$.
$\text{A \to R}$
(B) Given,& \left[\overset{→}{a}\overset{→}{b}\overset{→}{c}\right]$$=5 . . .$$\left(1\right)$</p><p>$V = \l \left[3 \left(\overset{→}{a} + \overset{→}{b}\right)\left(\overset{→}{b} + \overset{→}{c}\right)2 \left(\overset{→}{c} + \overset{→}{a}\right)\right]$</p><p>$⇒ V = 6 \left[ \left(\overset{→}{a} + \overset{→}{b}\right) \left(\overset{→}{b} + \overset{→}{c}\right)\left(\overset{→}{c} + \overset{→}{a}\right)\right] . . .\left(2\right)$</p><p>Using results,$\left[\left(\overset{→}{a} + \overset{→}{b}\right)\left(\overset{→}{b} + \overset{→}{c}\right)\left(\overset{→}{c} + \overset{→}{a}\right)\right] =2 \left[\left(\overset{→}{a}\right)\left(\overset{→}{b}\right)\left(\overset{→}{c}\right)\right] . . . \left(3\right)
$⇒ V = 6 \times2$$\left[\overset{→}{a}\overset{→}{b}\overset{→}{c}\right]$$</p><p>$⇒ V = 12 \left[\overset{→}{a}\overset{→}{b}\overset{→}{c}\right]$<br />Putting$ the $value \left[\overset{→}{a}\overset{→}{b}\overset{→}{c}\right]$ from equation \left(1\right)$, we get the volume of the parallelepiped ,$

&$⇒ V = 12 \times5 = 60$.
\text{B⟶S}
(C)) Given,Areaof triangle whose adjacents sides are&\overset{→}{a}and\overset{→}{b},

$⇒ ∆ =$\dfrac{1}{2}\left|\overset{→}{a} \times \overset{→}{b}\right|$=20 . . . \left(1\right)$
{Δ}_{1} = \dfrac{1}{2}\left|\left( 2 \overset{→}{a} + 3 \overset{→}{b}\right)\times \left(\overset{→}{a} - \overset{→}{b}\right)\right|<br />⇒ {∆}_{1} = \dfrac{1}{2}\left|-$-2 \left(\overset{→}{a} \times \overset{→}{b}\right)$ - -3 \left(\overset{→}{a} \times \overset{→}{b}\right)\right|<br />⇒ {∆}_{1} = 5 \left(\dfrac{1}{2}\left|\overset{→}{a} \times \overset{→}{b}\right|\right)
Using the results from equation& \left(1\right)

$⇒ {∆}_{1} = 5 \times20 = 100$
$\text{C⟶P}$
(D) Given,$The$area of the parallelogram with adjacent sides determined by vectors \left(\overset{→}{a}\right)and& \left(\overset{→}{b}\right)

\left|\overset{→}{a} \times \overset{→}{b}\right|$=30$. . . \left(1\right)$<br /> The area of the parallelogram with adjacent sides determined by vectors &\left(\overset{→}{a} + \overset{→}{b}\right)and&$\left(\overset{→}{a}\right)$ & & & & & & & & & & & & & & & & & & $</p><p>$⇒ AreaA = \left|\left(\overset{→}{a} + \overset{→}{b}\right) \times \left(\overset{→}{a}\right)\right| = \left|\overset{→}{b} \times \overset{→}{a}\right|, Because& \left(\overset{→}{a} \times \overset{→}{a}\right)$=0$

$⇒ AreaA = \left|\overset{→}{b} \times \overset{→}{a}\right|$

Using results$from$$\left(1\right)$,we get$</p><p>$⇒ AreaA = \left|\overset{→}{b} \times \overset{→}{a}\right| =$30$
\text{D⟶Q}