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

Solution :
$$\eqalign{ & {\text{Here }}\left( {\overrightarrow p .\overrightarrow p } \right)\left( {\overrightarrow x - \overrightarrow q } \right) - \left\{ {\overrightarrow p .\left( {\overrightarrow x - \overrightarrow q } \right)} \right\}\overrightarrow p + \left( {\overrightarrow q .\overrightarrow q } \right)\left( {\overrightarrow x - \overrightarrow r } \right) - \left\{ {\overrightarrow q .\left( {\overrightarrow x - \overrightarrow r } \right)} \right\}\overrightarrow q + \left( {\overrightarrow r .\overrightarrow r } \right)\left( {\overrightarrow x - \overrightarrow p } \right) - \left\{ {\overrightarrow r .\left( {\overrightarrow x - \overrightarrow p } \right)} \right\}\overrightarrow r = 0 \cr & {\text{or }}{\lambda ^2}\left( {\overrightarrow x - \overrightarrow q + \overrightarrow x - \overrightarrow r + \overrightarrow x - \overrightarrow p } \right) = \left\{ {\overrightarrow p .\left( {\overrightarrow x - \overrightarrow q } \right)} \right\}\overrightarrow p + \left\{ {\overrightarrow q .\left( {\overrightarrow x - \overrightarrow r } \right)} \right\}\overrightarrow q + \left\{ {\overrightarrow r .\left( {\overrightarrow x - \overrightarrow p } \right)} \right\}\overrightarrow r , \cr & {\text{where }}\left| {\overrightarrow p } \right| = \left| {\overrightarrow q } \right| = \left| {\overrightarrow r } \right| = \lambda \cr & {\text{or }}{\lambda ^2}\left\{ {3\overrightarrow x - \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right)} \right\} = \left( {\overrightarrow p .\overrightarrow x } \right)\overrightarrow p + \left( {\overrightarrow q .\overrightarrow x } \right)\overrightarrow q + \left( {\overrightarrow r .\overrightarrow x } \right)\overrightarrow r , \cr & {\text{because }}\overrightarrow p ,\,\overrightarrow q ,\,\overrightarrow r {\text{ are mutually perpendicular}}{\text{.}} \cr & {\text{Let }}\overrightarrow x = \alpha \overrightarrow p + \beta \overrightarrow q + \gamma \overrightarrow r {\text{. Then}} \cr & \,\,\,\,\,\,\,\,\,\,\,\overrightarrow p .\overrightarrow x = \alpha {\left| {\overrightarrow p } \right|^2} = \alpha {\lambda ^2},\,\overrightarrow q .\overrightarrow x = \beta {\left| {\overrightarrow q } \right|^2} = \beta {\lambda ^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\overrightarrow r .\overrightarrow x = \gamma {\left| {\overrightarrow r } \right|^2} = \gamma {\lambda ^2} \cr & \therefore \,{\lambda ^2}\left\{ {3\overrightarrow x - \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right)} \right\} = {\lambda ^2}\overrightarrow x \,\,\,\,{\text{or }}3\overrightarrow x - \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right) = \overrightarrow x \cr} $$