Let → a , → b and → c be three non - coplanar unit vectors such...

\left|\overset{\to}{a}\right| = \left|\overset{\to}{b}\right| = \left|\overset{\to}{c}\right| = 1$<br />$\overset{\to}{a} \times \overset{\to}{b} + \overset{\to}{b} \times \overset{\to}{c} = p\overset{\to}{a} + q\overset{\to}{b} + r\overset{\to}{c}$<br />$\overset{\to}{a} . \left(\overset{\to}{b} \times \overset{\to}{c}\right) = p + q\left(\overset{\to}{a} . \overset{\to}{b}\right) + r\left(\overset{\to}{a} . \overset{\to}{c}\right)$<br />And \left[\overset{\to}{a}\overset{\to}{b}\overset{\to}{c}\right] = \dfrac{1}{\sqrt{2}}$
$p + \dfrac{q}{2} + \dfrac{r}{2} = \left[\overset{\to}{a}\overset{\to}{b}\overset{\to}{c}\right]$ .......(i)
$\dfrac{p}{2} + q + \dfrac{r}{2} = 0$ ......(ii)
$\dfrac{p}{2} + \dfrac{q}{2} + r = \left[\overset{\to}{a}\overset{\to}{b}\overset{\to}{c}\right]$ ......(iii)
$⇒$ $p = r = - q$
$\dfrac{{p}^{2} + 2{q}^{2} + {r}^{2}}{{q}^{2}} = 4$