Question

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

Answer 1

Answer: 108

Equation reduces to $(4\vec{a}+3\vec{b})\cdot \vec{c}=7\vec{a}\cdot \vec{b}$
But $\vec{a}+\vec{b}+\vec{c}=0$
So, $(4\vec{a}+3\vec{b})\cdot (\vec{a}+\vec{b})=-7\vec{a}\cdot \vec{b}$
$\vec{a}\cdot \vec{b}=-6$
$|\vec{a}\times \vec{b}{|}^2=108$