If G(bar{g}),H(bar{h}),and P(bar{p}) are resp | vedclass

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(C) In any triangle,the centroid $G$ divides the line segment joining the orthocenter $H$ and the circumcentre $P$ in the ratio $2:1$. Thus,by the sec

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$2, 1, -3$

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(C) In any triangle,the centroid $G$ divides the line segment joining the orthocenter $H$ and the circumcentre $P$ in the ratio $2:1$. Thus,by the section formula,the position vector of the centroid is given by $\bar{g} = \frac{1 \cdot \bar{h} + 2 \cdot \bar{p}}{1 + 2}$. This simplifies to $3 \bar{g} = \bar{h} + 2 \bar{p}$,which can be rewritten as $2 \bar{p} + 1 \bar{h} - 3 \bar{g} = \overline{0}$. Comparing this with the given equation $x \bar{p} + y \bar{h} + z \bar{g} = \overline{0}$,we get $x = 2$,$y = 1$,and $z = -3$.

If $G(\bar{g})$,$H(\bar{h})$,and $P(\bar{p})$ are respectively the centroid,orthocenter,and circumcentre of a triangle and $x \bar{p} + y \bar{h} + z \bar{g} = \overline{0}$,then $x, y, z$ are respectively:

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