Question

In a triangle $P Q R$, let $\vec{a}=\overrightarrow{Q R}, \vec{b}=\overrightarrow{R P}$ and $\vec{c}=\overrightarrow{P Q}$.
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

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, $(4 \vec{a}+3 \vec{b}) \cdot(\vec{a}+\vec{b})=-7 \vec{a} \cdot \vec{b}$
$\vec{a} \cdot \vec{b}=-6$
$|\vec{a} \times \vec{b}|^{2}=108$