Let $$\left|\vec{p}\right|=\left|\vec{q}\right|=\left|\vec{p}\right|=a\;,$$ Then Given $$\vec{p}\cdot \vec{q}=\vec{q}\cdot \vec{r}=\vec{r}\cdot \vec{p} = 0.$$
And Given $$\begin{aligned} \vec p \times ((\vec x - \vec q) \times \vec p)+\vec q \times ((\vec x - \vec r) \times \vec q)+\vec r \times ((\vec x - \vec p) \times \vec r)=0\end{aligned}$$
Using Vector Triple Product Property, We Get
$$\Rightarrow \displaystyle \vec{p}\times \left(\vec{x}\times \vec{p}-\vec{q}\times \vec{p}\right)+\vec{q}\times \left(\vec{x}\times \vec{q}-\vec{r}\times \vec{q}\right)+\vec{r}\times \left(\vec{x}\times \vec{r}-\vec{p}\times \vec{r}\right)=0$$
$$\Rightarrow \displaystyle \vec{p}\times (\vec{x}\times \vec{p})- \vec{p}\times (\vec{q}\times \vec{p})+ \vec{q}\times (\vec{x}\times \vec{q}) - \vec{q}\times (\vec{r}\times \vec{q}) + \vec{r}\times (\vec{x}\times \vec{r}) - \vec{r}\times (\vec{p}\times \vec{r}) = 0$$.
$$\Rightarrow \displaystyle \left(\left|\vec{p}\right|^2+\left|\vec{q}\right|^2+\left|\vec{r}\right|^2\right)\vec{x}-\left(\vec{p}\cdot \vec{x}\right)\vec{p}- \left(\vec{q}\cdot \vec{x}\right)\vec{q}-\left(\vec{r}\cdot \vec{x}\right)\vec{r}-\left|\vec{p}\right|^2\vec{q}-\left|\vec{q}\right|^2\vec{r}-\left|\vec{r}\right|^2\vec{p}=0$$
Now above Given $$\vec{p}\;,\vec{q}\;,\vec{r}$$ are Three Perpendicular vector of same magnitude.
So We can write $$\vec{p}=a\vec{i}$$ and $$\vec{q}=a\vec{j}$$
and $$\vec{r}=a\vec{k}.$$ and $$\vec{x}=x_{1}\vec{i}+y_{1}\vec{j}+z_{1}\vec{k}$$
So Our equation convert into......
$$\displaystyle \Rightarrow 3a^2\vec{x}-a^2\left(x_{1}\vec{i}+y_{1}\vec{j}+z_{1}\vec{k}\right)-a^2(a\vec{i}+a\vec{j}+a\vec{k})=0$$
$$\displaystyle \Rightarrow 3a^2\vec{x}-a^2\vec{x}-a^2(\vec{p}+\vec{q}+\vec{r})=\vec{0}$$
So we get $$\displaystyle \vec{x} = \frac{1}{2}\left(\vec{p}+\vec{q}+\vec{r}\right)$$