Let $\left|\overrightarrow{a}\right|=2,\left|\overrightarrow{b}\right|=3$ and the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ be $\frac{\pi }{4}$. Then $\left|\right(\overrightarrow{a}+2\overrightarrow{b})\times (2\overrightarrow{a}-3\overrightarrow{b}){|}^2$ is equal to

Given,

$\left|\overrightarrow{a}\right|=2,\left|\overrightarrow{b}\right|=3$ and the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ be $\frac{\pi }{4}$,

Now solving,

$\left|\right(\overrightarrow{a}+2\overrightarrow{b})\times (2\overrightarrow{a}-3\overrightarrow{b}){|}^2$

$=|-3(\overrightarrow{a}\times \overrightarrow{b})+4(\overrightarrow{b}\times \overrightarrow{a}){|}^2$ $\left\{\text{as} \overrightarrow{a}\times \overrightarrow{a}=0 \& \overrightarrow{b}\times \overrightarrow{b}=0\right\}$

$=\left|7\right(\overrightarrow{b}\times \overrightarrow{a}){|}^2$ $\left\{\mathrm{as} \overrightarrow{b}\times \overrightarrow{a}=-\overrightarrow{a}\times \overrightarrow{b}\right\}$

$=49\left|a{|}^2\right|b{|}^2{\sin }^2\frac{\pi }{4}$

$=49\times 4\times 9\times \frac{1}{2}$

$=882$