$\overrightarrow{p},\overrightarrow{q}$ and $\overrightarrow{r}$ are three mutually perpendicular vectors of the same magnitude. If vector $\overrightarrow{x}$ satisfies the equation 

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

 

then $\overrightarrow{x}$ is given by

As $\overrightarrow{p},\overrightarrow{q}$ and $\overrightarrow{r}$ are three mutually perpendicular vectors of same magnitude, so let us consider 

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

Also let  $\overrightarrow{x}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$

Given that $\overrightarrow{x}$ satisfies the equation

$\begin{array}{rr} & \overrightarrow{p}\times \left[\right(\overrightarrow{x}-\overrightarrow{q})\times \overrightarrow{p}]+\overrightarrow{q}\times \left[\right(\overrightarrow{x}-\overrightarrow{r})\times \overrightarrow{q}]\\ & +\overrightarrow{r}\times \left[\right(\overrightarrow{x}-\overrightarrow{p})\times \overrightarrow{r}]=0\end{array}$

Now $\overrightarrow{p}\times \left[\right(\overrightarrow{x}-\overrightarrow{p})\times \overrightarrow{p}]=\overrightarrow{p}\times [\overrightarrow{x}\times \overrightarrow{p}-\overrightarrow{q}\times \overrightarrow{p}]$

$=\overrightarrow{p}\times (\overrightarrow{x}\times \overrightarrow{p})-\overrightarrow{p}\times (\overrightarrow{q}\times \overrightarrow{p})$

$=(\overrightarrow{p}\cdot \overrightarrow{p})\overrightarrow{x}-(\overrightarrow{p}\cdot \overrightarrow{x})\overrightarrow{p}-(\overrightarrow{p}\cdot \overrightarrow{p})\overrightarrow{q}+(\overrightarrow{p}\cdot \overrightarrow{q})\overrightarrow{p}$

$=a^2\overrightarrow{x}-a^2{x}_1\hat{i}-a^3\hat{j}+0$

Similarly, $\overrightarrow{q}\times \left[\right(\overrightarrow{x}-\overrightarrow{r})\times \overrightarrow{q}]=a^2\overrightarrow{x}-a^2{y}_1\hat{j}-a^3\hat{k}$

And $\overrightarrow{r}\times \left[\right(\overrightarrow{a}-\overrightarrow{p})\times \overrightarrow{r}]=a^3\overrightarrow{x}-a^2{z}_1\hat{k}-a^3\hat{i}$

Substituting values in the equation, we get

$3a^2\overrightarrow{x}-a^2\left(x_1\hat{i}+y_{1}j7+z_1\hat{k}\right)-a^2(a\hat{i}+a\hat{j}+a\hat{k})=0$

Or $3a^2\overrightarrow{x}-a^2\overrightarrow{x}-a^2(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})=\overrightarrow{0}$

Or $2a^2\overrightarrow{x}=(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})a^2$

Or $\overrightarrow{x}=\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$