In a triangle 𝑃𝑄𝑅, let $$\overrightarrow{a} = \overrightarrow{QR}, \overrightarrow{b} = \overrightarrow{RP}$$ and $$\overrightarrow{c} = \overrightarrow{PQ}$$. If

$$|\overrightarrow{a}| = 3, |\overrightarrow{b}| = 4$$ and $$\frac{\overrightarrow{a}\cdot(\overrightarrow{c}-\overrightarrow{b})}{\overrightarrow{c}\cdot(\overrightarrow{a}-\overrightarrow{b})} = \frac{|\overrightarrow{a}|}{|\overrightarrow{a}| + |\overrightarrow{b}|}$$,

then the value of $$\left|\overrightarrow{a} \times \overrightarrow{b}\right|^{2}$$ is ___________

In the triangle, choose position vectors so that

$$\overrightarrow{a}= \overrightarrow{QR}= \mathbf{r}-\mathbf{q},\; \overrightarrow{b}= \overrightarrow{RP}= \mathbf{p}-\mathbf{r},\; \overrightarrow{c}= \overrightarrow{PQ}= \mathbf{q}-\mathbf{p}.$$

Traversing the triangle once gives the vector identity

$$\overrightarrow{a}+ \overrightarrow{b}+ \overrightarrow{c}= \mathbf{0} \;\;\Longrightarrow\;\; \overrightarrow{c}= -(\overrightarrow{a}+ \overrightarrow{b}).$$

The given magnitudes are $$|\overrightarrow{a}|=3,\;|\overrightarrow{b}|=4.$$

Start with the ratio supplied in the question:

$$\frac{\overrightarrow{a}\!\cdot\!(\overrightarrow{c}-\overrightarrow{b})} {\overrightarrow{c}\!\cdot\!(\overrightarrow{a}-\overrightarrow{b})} =\frac{|\overrightarrow{a}|}{|\overrightarrow{a}|+|\overrightarrow{b}|} =\frac{3}{3+4}=\frac{3}{7}.$$

Compute the two scalar products in the ratio using $$\overrightarrow{c}=-(\overrightarrow{a}+\overrightarrow{b}).$$

1. $$\overrightarrow{c}-\overrightarrow{b} =-(\overrightarrow{a}+\overrightarrow{b})-\overrightarrow{b} =-\overrightarrow{a}-2\overrightarrow{b}.$$
Hence
$$\overrightarrow{a}\!\cdot\!(\overrightarrow{c}-\overrightarrow{b}) =\overrightarrow{a}\!\cdot\!(-\overrightarrow{a}-2\overrightarrow{b}) =-|\overrightarrow{a}|^{2}-2\,\overrightarrow{a}\!\cdot\!\overrightarrow{b}.$$

2. $$\overrightarrow{a}-\overrightarrow{b}$$ is unchanged, so
$$\overrightarrow{c}\!\cdot\!(\overrightarrow{a}-\overrightarrow{b}) =[-(\overrightarrow{a}+\overrightarrow{b})]\!\cdot\!(\overrightarrow{a}-\overrightarrow{b}) =-|\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}.$$

Insert these into the ratio:

$$\frac{-|\overrightarrow{a}|^{2}-2\,\overrightarrow{a}\!\cdot\!\overrightarrow{b}} {\,|\overrightarrow{b}|^{2}-|\overrightarrow{a}|^{2}} =\frac{3}{7}.$$

With $$|\overrightarrow{a}|^{2}=9,\;|\overrightarrow{b}|^{2}=16,$$ the denominator is $$16-9=7,$$ so

$$-9-2\,\overrightarrow{a}\!\cdot\!\overrightarrow{b}=3 \;\;\Longrightarrow\;\; \overrightarrow{a}\!\cdot\!\overrightarrow{b}=-6.$$

The square of the magnitude of the cross‐product follows from the identity

$$|\overrightarrow{a}\times\overrightarrow{b}|^{2} =|\overrightarrow{a}|^{2}|\overrightarrow{b}|^{2} -(\overrightarrow{a}\!\cdot\!\overrightarrow{b})^{2}.$$

Therefore

$$|\overrightarrow{a}\times\overrightarrow{b}|^{2} =9\!\times\!16-(-6)^{2} =144-36 =108.$$

Hence the required value is 108.