If Let $\mathrm{△}PQR$ be a triangle. Let $\overrightarrow{a}=\overrightarrow{QR}, \overrightarrow{b}=\overrightarrow{RP}$ and $\overrightarrow{c}=\overrightarrow{PQ}$. If $\left|\overrightarrow{a}\right|=12,\left|\overrightarrow{b}\right|=4\sqrt{3}$ and $\overrightarrow{b}\cdot \overrightarrow{c}=24$, then which of the following is (are) true?

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\Rightarrow \overrightarrow{b}+\overrightarrow{c}=-\overrightarrow{a}$

$\Rightarrow |\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+2\overrightarrow{b}\cdot \overrightarrow{c}=|\overrightarrow{a}{|}^2$

$\Rightarrow 48+|\overrightarrow{c}{|}^2+48=144\Rightarrow |\overrightarrow{c}{|}^2=48\Rightarrow \left|\overrightarrow{c}\right|=4\sqrt{3}$

$\therefore \frac{|\overrightarrow{c}{|}^2}{2}-\left|\overrightarrow{a}\right|=24-12=12\Rightarrow \frac{|\overrightarrow{c}{|}^2}{2}+\left|\overrightarrow{a}\right|=36$

Further, $\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}\Rightarrow |\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+2\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{c}{|}^2$

$\Rightarrow 144+48+2\overrightarrow{a}\cdot \overrightarrow{b}=-48\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=-72$

$\because \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\Rightarrow \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}=0$

$\therefore |\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{c}\times \overrightarrow{a}|=2|\overrightarrow{a}\times \overrightarrow{b}|$

$=2\sqrt{a^2{b}^2-(\overrightarrow{a}\cdot \overrightarrow{b}{)}^2}=2\sqrt{\left(144\right)\left(48\right)-(72{)}^2}=48\sqrt{3}$