In a triangle PQR ,let a=QR,b=RP and c=PQ. If $\left|a\right|=3,\left|b\right|=4and\frac{a.\left(c-b\right)}{c.\left(a-b\right)}=\frac{\left|a\right|}{\left|a\right|+\left|b\right|}$ then the value of${\left|a\times b\right|}^2$  is ______________
 

It is given that, in a $\Delta PQR$ 
   $PQ=c,$$QR=a$, RP=b 
   $a+b+c=0$
   $\therefore |a{|}^2+a.b+a.c=0$……. (i)
 and   $a.b+|b{|}^2+c.b=0$   …….(ii)
and also given that  $\left|a\right|=3,\left|b\right|=4$
and  $$$\frac{a.c-a.b}{c.a-c.b}=\frac{\left|a\right|}{\left|a\right|+\left|b\right|}=\frac{3}{7}$
  $\Rightarrow 7\left[\left(a.c\right)-\left(a.b\right)\right]=3\left[\left(c.a\right)-\left(c.b\right)\right]$
from Eqs. (i) and (ii) on putting the values of a .c and c . b, we get
  
  $\Rightarrow 7\left[-9-(a.b)-(a.b)\right]=3\left[-9-(a.b)+16+(a.b)\right]$
  $\Rightarrow 7\left[-9-2(a.b)\right]=3\times 7$
  $\Rightarrow a.b=-6$
      $\because |a\times b{|}^2={\left(\left|a\right|\left|b\right|\right)}^2-{(a.b)}^2$
 $=144-36$                       $=108$