Let $\overrightarrow{p},\overrightarrow{q},\overrightarrow{r}$ be three mutually perpendicular vectors 

of the same magnitude. If a vector $\overrightarrow{x}$ satisfies the equation

$\left\{\overrightarrow{p}\times \left((\overrightarrow{x}-\overrightarrow{q})\times \overrightarrow{p}\right)\right\}+\left\{\overrightarrow{q}\times \left((\overrightarrow{x}-\overrightarrow{r})\times \overrightarrow{q}\right)\right\}+\left\{\overrightarrow{r}\times \left(\left(\overrightarrow{x}-\overrightarrow{p}\right)\times \overrightarrow{r}\right)\right\}=\overrightarrow{0}$

then $\overrightarrow{x}$ is given by

We have,

$\begin{array}{l}\overrightarrow{p}\times \left\{\right(\overrightarrow{x}-\overrightarrow{q})\times \overrightarrow{p}\}+\overrightarrow{q}\times \left\{\right(\overrightarrow{x}-\overrightarrow{r})\times \overrightarrow{q}\}+\overrightarrow{r}\times \{\overrightarrow{x}-\overrightarrow{p})\times \overrightarrow{r}\}=\overrightarrow{0}\\ \Rightarrow (\overrightarrow{p}\cdot \overrightarrow{p})(\overrightarrow{x}-\overrightarrow{q})-\{\overrightarrow{p}\cdot (\overrightarrow{x}-q\overrightarrow{)p})\mid +(\overrightarrow{q}\cdot q\left)\right(\overrightarrow{x}-\overrightarrow{r})-\{(\overrightarrow{q}\cdot (\overrightarrow{x}-\overrightarrow{r}\left)\right\}\overrightarrow{q}\\ +(\overrightarrow{r}\cdot \overrightarrow{r})(\overrightarrow{x}-\overrightarrow{p})-\{\overrightarrow{r}\cdot (\overrightarrow{x}-\overrightarrow{p}\left)\right\}\overrightarrow{r}=\overrightarrow{0}\\ \Rightarrow \left(|\overrightarrow{p}{|}^2+|\overrightarrow{q}{|}^2+|\overrightarrow{r}{|}^2\right)\overrightarrow{x}-\left(|\overrightarrow{p}{|}^2\overrightarrow{q}+|\overrightarrow{q}{|}^2\overrightarrow{r}+|\overrightarrow{r}{|}^2\overrightarrow{p}\right)\\ -\{\overrightarrow{p}\cdot \overrightarrow{x})\overrightarrow{p}+(\overrightarrow{q}\cdot \overrightarrow{x})\overrightarrow{q}+(\overrightarrow{r}\cdot \overrightarrow{x})\overrightarrow{r}\}=0\\ 3{\lambda }^2\overrightarrow{x}-{\lambda }^2(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})-\left\{\right(\overrightarrow{p}\cdot \overrightarrow{x})\overrightarrow{p}+(\overrightarrow{q}\cdot \overrightarrow{x})\overrightarrow{q}+(\overrightarrow{r}\cdot \overrightarrow{x}\left)\overrightarrow{r}\right\}=0\end{array}$

where $\lambda =\left|\overrightarrow{p}\right|=\left|\overrightarrow{q}\right|=\left|\overrightarrow{r}\right|$

$\begin{array}{l}\Rightarrow 3{\lambda }^2\overrightarrow{x}-{\lambda }^2(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})-{\lambda }^2\overrightarrow{x}=\overrightarrow{0}\\ \left[\because \overrightarrow{x}=\frac{(\overrightarrow{p}\cdot \overrightarrow{x})\overrightarrow{p}}{|\overrightarrow{p}{|}^2}+\frac{(\overrightarrow{q}\cdot \overrightarrow{x})\overrightarrow{q}}{|\overrightarrow{q}{|}^2}+\frac{(\overrightarrow{r}\cdot \overrightarrow{x})\overrightarrow{r}}{|\overrightarrow{r}{|}^2}\right]\\ \Rightarrow \overrightarrow{x}=\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})\end{array}$