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

Solution :
Let $$\left| {\overrightarrow p } \right| = \left| {\overrightarrow q } \right| = \left| {\overrightarrow r } \right| = k$$
Let $$\hat p,\,\hat q,\,\hat r$$   be unit vectors along $$\overrightarrow p ,\,\overrightarrow q ,\,\overrightarrow r $$   respectively. Clearly $$\hat p,\,\hat q,\,\hat r$$   are mutually perpendicular vectors, so any vector $$\overrightarrow x $$ can be written as $${a_1}\hat p + {a_2}\hat q + {a_3}\hat r.$$
$$\eqalign{ & \therefore \,\overrightarrow p \times \left\{ {\left( {\overrightarrow x - \overrightarrow q } \right) \times \overrightarrow p } \right\} = \left( {\overrightarrow p .\overrightarrow p } \right)\left( {\overrightarrow x - \overrightarrow q } \right) - \left\{ {\overrightarrow {p.} \left( {\overrightarrow x - \overrightarrow q } \right)} \right\}\overrightarrow p \cr & = {k^2}\left( {\overrightarrow x - \overrightarrow q } \right) - \left( {\overrightarrow p .\overrightarrow x } \right)\overrightarrow p \,\,\,\,\,\,\,\,\,\left[ {\because \,\overrightarrow p .\overrightarrow q = 0} \right] \cr & = {k^2}\left( {\overrightarrow x - \overrightarrow q } \right) - k\hat p.\left( {{a_1}\hat p + {a_2}\hat q + {a_3}\hat r} \right)k\hat p \cr & = {k^2}\left( {\overrightarrow x - \overrightarrow q - {a_1}\hat p} \right) \cr & {\text{Similarly, }}\overrightarrow q \times \left\{ {\left( {\overrightarrow x - \overrightarrow r } \right) \times \overrightarrow q } \right\} = {k^2}\left( {\overrightarrow x - \overrightarrow r - {a_2}\hat q} \right) \cr & {\text{and }}\overrightarrow r \times \left\{ {\left( {\overrightarrow x - \overrightarrow p } \right) \times \overrightarrow r } \right\} = {k^2}\left( {\overrightarrow x - \overrightarrow p - {a_3}\hat r} \right) \cr} $$
According to the given condition
$$\eqalign{ & {k^2}\left( {\overrightarrow x - \overrightarrow q - {a_1}\hat p + \overrightarrow x - \overrightarrow r - {a_2}\hat q + \overrightarrow x - \overrightarrow p - {a_3}\hat r} \right) = 0 \cr & \Rightarrow {k^2}\left\{ {3\overrightarrow x - \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right) - \left( {{a_1}\hat p + {a_2}\hat q + {a_3}\hat r} \right)} \right\} = 0 \cr & \Rightarrow {k^2}\left[ {2\overrightarrow x - \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right)} \right] = \overrightarrow 0 \cr & \Rightarrow \overrightarrow x = \frac{1}{2}\overrightarrow {} \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right)\,\,\,\,\,\left[ {\because \,k \ne 0} \right] \cr} $$