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Let $ABC $ be a triangle whose circumcentre is at $P$. If the position vectors of $A, B, C$ and $P$ are $\overrightarrow{a} , \overrightarrow{b}, \overrightarrow{c}$ and $\frac{\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}}{4}$ respectively, then the position vector of the orthocentre of this triangle, is :

$\begin{array}{1 1}(1)\; \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \\ (2)\; -\begin{bmatrix} \frac{\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} }{2} \end{bmatrix} \\ (3)\; \overrightarrow{0} \\ (4)\; \frac{(\overrightarrow{a}+ \overrightarrow{b}+\overrightarrow{c} )}{2} \end{array} $

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