Question
Let $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ be any four points in space. Prove that
$|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|=4$ (area of $\triangle A B C$ )
$|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|=4$ (area of $\triangle A B C$ )
Consider $|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|$
$\begin{aligned}=(\bar{b}-\bar{a}) \times(\bar{d}-\bar{c})+(\bar{c}-\bar{b}) \times(\bar{d}-\bar{a})+(\bar{a}-\bar{c}) \times(\bar{d}-\bar{b}) \\ =\bar{b} \times(\bar{d}-\bar{c})-\bar{a} \times(\bar{d}-\bar{c})+\bar{c} \times(\bar{d}-\bar{a})-\bar{b} \times(\bar{d}-\bar{a})+ \\ \bar{a} \times(\bar{d}-\bar{b})-\bar{c} \times(\bar{d}-\bar{b})\end{aligned}$
$\begin{array}{r}=\bar{b} \times \bar{d}-\bar{b} \times \bar{c}-\bar{a} \times \bar{d}+\bar{a} \times \bar{c}+\bar{c} \times \bar{d}-\bar{c} \times \bar{a}-\bar{b} \times \bar{d}+ \\ \bar{b} \times \bar{a}+\bar{a} \times \bar{d}-\bar{a} \times \bar{b}-\bar{c} \times \bar{d}+\bar{c} \times \bar{b} \\ =\bar{b} \times \bar{d}-\bar{b} \times \bar{c}-\bar{a} \times \bar{d}-\bar{c} \times \bar{a}+\bar{c} \times \bar{d}-\bar{c} \times \bar{a}-\bar{b} \times \bar{d}- \\ \bar{a} \times \bar{b}+\bar{a} \times \bar{d}-\bar{a} \times \bar{b}-\bar{c} \times \bar{d}-\bar{b} \times \bar{c} \\ \ldots[\because \bar{p} \times \bar{q}=-\bar{q} \times \bar{p}]\end{array}$
$-2(\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a})$
$\begin{aligned} & \therefore|\overline{\mathrm{AB}} \times \overline{\mathrm{CD}}+\overline{\mathrm{BC}} \times \overline{\mathrm{AD}}+\overline{\mathrm{CA}} \times \overline{\mathrm{BD}}| \\ & =|-2(\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a})| \\ & =4\left[\frac{1}{2}|\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a}|\right]\end{aligned}$
$=4($ area of $\triangle A B C)$
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