Let $\overrightarrow{p},\overrightarrow{q},\overrightarrow{r}$ be three mutually perpendicular vectors of same magnitude. If a vector $\overrightarrow{x}$satisfies the equation $\overrightarrow{p}\times \left(\right(\overrightarrow{x}-\overrightarrow{q})\times \overrightarrow{p})+\overrightarrow{q}\times \left(\right(\overrightarrow{x}-\overrightarrow{r})\times \overrightarrow{q})+\overrightarrow{r}\times \left(\right(\overrightarrow{x}-\overrightarrow{p})\times \overrightarrow{r})=0$ then $\overrightarrow{x}=$

$\overrightarrow{p}=p\widehat{i},\overrightarrow{q}=p\widehat{j},\overrightarrow{r}=p\widehat{k}$

$\overrightarrow{p}\times \left(\right(\overrightarrow{x}-\overrightarrow{q})\times \overrightarrow{p})=p^2\widehat{i}\times \left(\right(\overrightarrow{x}-p\widehat{j})\times \widehat{i})$.

$=p^2[\widehat{i}\cdot \widehat{i}(\overrightarrow{x}-p\cdot \widehat{j})-\widehat{i}\cdot \overrightarrow{x}\widehat{i}]=p^2[\overrightarrow{x}-p\widehat{j}-(\widehat{i}\cdot \overrightarrow{x}\left)\widehat{i}\right]$

Now, $\mathrm{\Sigma }\overrightarrow{p}\times \left(\right(\overrightarrow{x}-\overrightarrow{q})\times \overrightarrow{p})=0$

$\Rightarrow \mathrm{\Sigma }(\overrightarrow{x}-p\cdot \widehat{j}-(\widehat{i}\cdot \overrightarrow{x}\left)\widehat{i}\right)=0$

$\Rightarrow 3\overrightarrow{x}-p(\widehat{i}+\widehat{j}+\widehat{k})-\overrightarrow{x}=0$

$\therefore \overrightarrow{x}=\frac{p}{2}(\widehat{i}+\widehat{j}+\widehat{k})=\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$