In a triangle PQR , let → a = → QR , → b = → RP and → c = → PQ . If...

$\overset{\to}{a} + \overset{\to}{b} + \overset{\to}{c} = 0$

$∵ \dfrac{\overset{\to}{a} . \left(\overset{\to}{c} - \overset{\to}{b}\right)}{\overset{\to}{c} . \left(\overset{\to}{a} - \overset{\to}{b}\right)} = \dfrac{\left|\overset{\to}{a}\right|}{\left|\overset{\to}{a}\right| + \left|\overset{\to}{b}\right|}$

$⇒ \dfrac{- \left(\overset{\to}{b} + \overset{\to}{c}\right) ⋅ \left(\overset{\to}{c} - \overset{\to}{b}\right)}{- \left(\overset{\to}{a} + \overset{\to}{b}\right) ⋅ \left(\overset{\to}{a} - \overset{\to}{b}\right)} = \dfrac{\left|\overset{\to}{a}\right|}{\left|\overset{\to}{a}\right| + \left|\overset{\to}{b}\right|}$

$⇒ \dfrac{{\left|\overset{\to}{c}\right|}^{2} - {\left|\overset{\to}{b}\right|}^{2}}{{\left|\overset{\to}{a}\right|}^{2} - {\left|\overset{\to}{b}\right|}^{2}} = \dfrac{3}{3 + 4} = \dfrac{3}{7}$

$⇒ {\left|\overset{\to}{c}\right|}^{2} = 13$

Now, $\overset{\to}{a} + \overset{\to}{b} = - \overset{\to}{c} ⇒ {\left|\overset{\to}{a} + \overset{\to}{b}\right|}^{2} = {\left|- \overset{\to}{c}\right|}^{2}$

$⇒ {\left|\overset{\to}{a}\right|}^{2} + {\left|\overset{\to}{b}\right|}^{2} + 2\overset{\to}{a} . \overset{\to}{b} = {\left|\overset{\to}{c}\right|}^{2}$

$⇒ 9 + 16 + 2\left(\overset{\to}{a} . \overset{\to}{b}\right) = 13$

$⇒ \overset{\to}{a} \cdot \overset{\to}{b} = - 6$

Now, ${\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} + {\left(\overset{\to}{a} . \overset{\to}{b}\right)}^{2} = {\left|\overset{\to}{a}\right|}^{2} . {\left|\overset{\to}{b}\right|}^{2}$

$⇒ {\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} + 36 = \left(9\right)\left(16\right) ⇒ {\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} = 108$