Vector - triple cross product

Question:

If $$\vec p \times (\vec u \times \vec p) + \vec q \times (\vec v \times \vec q) + \vec r \times (\vec w \times \vec r) = 0$$ Where $\vec p, \vec q, \vec r$ are mutually perpendicular vectors with the same magnitude and: $$\vec u = \vec X - \vec q$$ $$\vec v = \vec X - \vec r$$ $$\vec w = \vec X - \vec p$$ Then $\vec X$ can be expressed as?

Not sure how I should proceed with this question. As it is given that $p,q,r$ are perpendicular vectors with the same magnitude, I tried to suppose they are unit vectors along the three axes, however, that didn't help. I attempted to use the fact that: $$\vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c)\vec b - (\vec a \cdot \vec b) \vec c$$

However, I was unable to proceed from there as well. What should I do?

I think this works: we have that

$$0 = \vec p \times (\vec X \times \vec p - \vec q \times \vec p) + \vec q \times (\vec X \times \vec q - \vec r \times \vec q) + \vec r \times (\vec X \times \vec r - \vec p \times \vec r)$$ instead of $\vec p, \vec q, \vec r$, lets call them $p_1 , p_2 , p_3$; in fact, we are going to be using cyclical notation, so lets just say that $p_{3k+1} = \vec p$, $p_{3k+2} = \vec q$, and $p_{3k} = \vec r$ for any $k$. Then, $$\sum _i p_i \times ( p_{i+1} \times p_i ) = \sum _i p_i \times ( X \times p_i ).$$ Now, since three mutually orthogonal vectors are a frame for $\mathbb{R}^3$, we know that $p_i \times p_{i+1} = \left( \pm \frac{|p_i||p_{i+1}|}{|p_{i+2}|} \right) p_{i+2}$. So, the sum on the left becomes $$\sum _i p_i \times ( p_{i+1} \times p_i ) = \sum _i p_i \times ( \left( \mp \frac{|p_i||p_{i+1}|}{|p_{i+2}|} \right) p_{i+2} ) = \sum _i \left( \mp \frac{|p_i||p_{i+1}|}{|p_{i+2}|} \right) p_i \times ( p_{i+2} ) = \sum _i \left( \mp \frac{|p_i||p_{i+1}|}{|p_{i+2}|} \right) p_i \times ( p_{i+{(3=0)} - 1} ) = \sum _i \left( \mp \frac{|p_i||p_{i+1}|}{|p_{i+2}|} \right) \left( \mp \frac{|p_i||p_{i-1}|}{|p_{i+1}|} \right) p_{i+1} = \sum _i |p_i|^2 p_{i+1},$$ that is to say, merely $$| \vec p |^2 \vec q + | \vec q |^2 \vec r + | \vec r |^2 \vec p.$$ Then, by the lagrange (?) formula, the right hand sum becomes $$\sum _i p_i \times ( X \times p_i ) = \sum_i | p_i |^2 X - \sum_i ( p_i \cdot X ) p_i.$$ So, $$X = \frac{ \sum _i |p_i|^2 p_{i+1} + \sum_i ( p_i \cdot X ) p_i }{ \sum_i | p_i |^2 } = \frac{ \sum _i (|p_{i+2}|^2 + p_i \cdot X ) p_{i} }{ \sum_i | p_i |^2 } .$$ Unfortunately, off the top of my head, I am not sure how to go much farther...

EDIT: I thought about how to go further

Since the $p_i$ are a frame, we can write down the coordinate for $X$ relative to that frame: $X = x_{p_1} p_1 + x_{p_2} p_2 + x_{p_3} p_3 = \sum_i x_{p_i} p_i$. Then, we must have from the last equation that $$x_{p_i} = \frac{ (|p_{i+2}|^2 + x_{p_i} )}{ \sum_i | p_i |^2 },$$ so that $$x_{p_i} = \frac{|p_{i+2}|^2}{\sum_i | p_i |^2 - 1}.$$ hence, $X = \frac{1}{ |p|^2 + |q|^2 + |r|^2 - 1 } ( |r|^2 , |p|^2 , |q|^2 )$.

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 $$\displaystyle \vec{x} = \frac{1}{2}\left(\vec{p}+\vec{q}+\vec{r}\right)$$