Let → a , → b , → c be three vectors such that | → a | = 31 , 4| →...

Given,

$\left|\overset{\to}{a}\right| = \sqrt{31}, 4\left|\overset{\to}{b}\right| = \left|\overset{\to}{c}\right| = 2$, $2\left(\overset{\to}{a} \times \overset{\to}{b}\right) = 3\left(\overset{\to}{c} \times \overset{\to}{a}\right)$ and angle between $\overset{\to}{b} \text{ and } \overset{\to}{c}$ is given as $\dfrac{2\pi}{3}$

Now solving, $3\left(\overset{\to}{c} \times \overset{\to}{a}\right) + 2\left(\overset{\to}{b} \times \overset{\to}{a}\right) = 0$

$⇒ \left(3\overset{\to}{c} \times 2\overset{\to}{b}\right) \times \overset{\to}{a} = 0$

Means \left(3\overset{\to}{c} \times 2\overset{\to}{b}\right) \text{ and } \overset{\to}{a}$are parallel vector,</p><p>So, let$3\overset{\to}{c} \times 2\overset{\to}{b} = \lambda\overset{\to}{a}$</p><p>Now squaring both sides we get,<br />$9{\left|\overset{\to}{c}\right|}^{2} + 4{\left|\overset{\to}{b}\right|}^{2} + 12\left(\overset{\to}{b} \cdot \overset{\to}{c}\right) = {\lambda}^{2}{\left|\overset{\to}{a}\right|}^{2}$</p><p>$⇒ 36 + 1 + 12 \times \dfrac{1}{2} \times 2\left(\cos \left(2\dfrac{\pi}{3}\right)\right) = {\lambda}^{2}\left(31\right)$</p><p>$⇒ {\lambda}^{2} = 1$</p><p>$⇒ \lambda = \pm 1$</p><p>Now putting the value of$\lambda$in \left(3\overset{\to}{c} \times 2\overset{\to}{b}\right) = \lambda\overset{\to}{a}$ we get,

$3\overset{\to}{c} + 2\overset{\to}{b} = \pm \overset{\to}{a} \ldots \left(1\right)$

Now taking dot product with $\overset{\to}{b}$ in above equation we get,

$3\left(\overset{\to}{b} \cdot \overset{\to}{c}\right) + 2\left(\overset{\to}{b} \cdot \overset{\to}{b}\right) = \pm \overset{\to}{a} \cdot \overset{\to}{b}$

$⇒ \overset{\to}{a} \cdot \overset{\to}{b} = \pm \left(- \dfrac{3}{2} + \dfrac{1}{2}\right) = \pm \left(- 1\right)$

$⇒ {\left(\overset{\to}{a} \cdot \overset{\to}{b}\right)}^{2} = 1$

Again taking $3\left(\overset{\to}{c} \times \overset{\to}{a}\right) = 2\left(\overset{\to}{a} \times \overset{\to}{b}\right)$ and sqauring both side,

$⇒ {\left(\overset{\to}{c} \times \overset{\to}{a}\right)}^{2} = \dfrac{4}{9}{\left(\overset{\to}{a} \times \overset{\to}{b}\right)}^{2}$

$⇒ {\left(\overset{\to}{c} \times \overset{\to}{a}\right)}^{2} = \dfrac{4}{9}\left[{\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2} - {\left(\overset{\to}{a} \cdot \overset{\to}{b}\right)}^{2}\right]$

$⇒ {\left(\overset{\to}{c} \times \overset{\to}{a}\right)}^{2} = \dfrac{4}{9}\left[\dfrac{31}{4} - \left(1\right)\right]$

$⇒ {\left(\overset{\to}{c} \times \overset{\to}{a}\right)}^{2} = \dfrac{4}{9} \times \dfrac{27}{4} = 3$

Hence, the value of ${\left(\dfrac{\overset{\to}{a} \times \overset{\to}{c}}{\overset{\to}{a} \cdot \overset{\to}{b}}\right)}^{2} = \dfrac{3}{1} = 3$.