Let $p, q, r$ be three mutually perpendicular  vectors of the same magnitude. If a vector $\overrightarrow{\mathbf{x}}$ satisfies the equation

$\overrightarrow{\mathbf{p}}\times \left(\right(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{q}})\times \overrightarrow{\mathbf{p}})+\overrightarrow{\mathbf{q}}\times \left(\right(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{r}})\times \overrightarrow{\mathbf{q}})+\overrightarrow{\mathbf{r}}\times \left(\right(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{p}})\times \overrightarrow{\mathbf{r}})=\overrightarrow{\mathbf{0}}$ then $\overrightarrow{x}$ is given by

$\text{ Let }\left|\mathbf{p}\right|=\left|\mathbf{q}\right|=\left|\mathbf{r}\right|=K.\text{ Let }\mathbf{a},\mathbf{b},\mathbf{c}$ be unit vectors along p, q, r respectively. Clearly $p,q,r$ are mutually perpendicular vectors, so any vector $\mathit{x}$ can be written as $a_1\mathbf{a}+a_2\mathbf{b}+a_3\mathbf{c}$

$\mathbf{p}\times \left(\right(\mathbf{x}-\mathbf{q})\times \mathbf{p})=(\mathbf{p}\cdot \mathbf{p})(\mathbf{x}-\mathbf{q})-(\mathbf{p}\cdot (\mathbf{x}-\mathbf{q}\left)\right)\mathbf{p}$

$\begin{array}{r}=K^2(\mathbf{x}-\mathbf{q})-(\mathbf{p}\cdot \mathbf{x})\mathbf{p}(\mathbf{p}\cdot \mathbf{q}=0)\\ =K^2(\mathbf{x}-\mathbf{q})-K\mathbf{a}\cdot \left(a_1\mathbf{a}+a_2\mathbf{b}+a_3\mathbf{c}\right)K\mathbf{a}\\ =K^2\left(\mathbf{x}-\mathbf{q}-a_1\mathbf{a}\right)\end{array}$

Similarly $\mathbf{q}\times \left(\right(\mathbf{x}-\mathbf{r})\times \mathbf{q})=K^2\left(\mathbf{x}-\mathbf{r}-a_2\mathbf{b}\right)$

and $\mathbf{r}\times \left(\right(\mathbf{x}-\mathbf{p})\times \mathbf{r})=K^2\left(\mathbf{x}-\mathbf{p}-a_3\mathbf{c}\right)$

According to the given condition 

$K^2\left(\mathbf{x}-\mathbf{q}-a_1\mathbf{a}+\mathbf{x}-\mathbf{r}-a_2\mathbf{b}+\mathbf{x}-\mathbf{p}-a_3\mathbf{c}\right)=0$

$\begin{array}{l}\Rightarrow \left\{3\mathbf{x}-(\mathbf{p}+\mathbf{q}+\mathbf{r})-\left(a_1\mathbf{a}+a_2\mathbf{b}+a_3\mathbf{c}\right)\right\}=0\\ \Rightarrow [2\mathbf{x}-(\mathbf{p}+\mathbf{q}+\mathbf{r}\left)\right]=\mathbf{0}\\ \Rightarrow x=(1/2)(\mathbf{p}+\mathbf{q}+\mathbf{r})\end{array}$