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(C) Let the origin be the circumcenter $O$ of the equilateral triangle $ABC.$ Let the position vectors of $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ and

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$2$

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(C) Let the origin be the circumcenter $O$ of the equilateral triangle $ABC.$ Let the position vectors of $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ and $P$ be $\vec{r}.$Since $O$ is the circumcenter,$|\vec{a}| = |\vec{b}| = |\vec{c}| = R,$ where $R$ is the circumradius.For an equilateral triangle of side $s=1,$ the circumradius $R = \frac{s}{\sqrt{3}} = \frac{1}{\sqrt{3}}.$Since $P$ is on the circumcircle,$|\vec{r}| = R = \frac{1}{\sqrt{3}}.$We need to calculate $S = |\vec{r}-\vec{a}|^2 + |\vec{r}-\vec{b}|^2 + |\vec{r}-\vec{c}|^2.$Expanding this,we get:$S = (\vec{r}-\vec{a}) \cdot (\vec{r}-\vec{a}) + (\vec{r}-\vec{b}) \cdot (\vec{r}-\vec{b}) + (\vec{r}-\vec{c}) \cdot (\vec{r}-\vec{c})$$S = 3|\vec{r}|^2 + (|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2) - 2\vec{r} \cdot (\vec{a}+\vec{b}+\vec{c}).$For an equilateral triangle centered at the origin,$\vec{a}+\vec{b}+\vec{c} = \vec{0}.$Also,$|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = R^2 = \frac{1}{3}$ and $|\vec{r}|^2 = R^2 = \frac{1}{3}.$Substituting these values:$S = 3(\frac{1}{3}) + (\frac{1}{3} + \frac{1}{3} + \frac{1}{3}) - 2\vec{r} \cdot \vec{0}$$S = 1 + 1 - 0 = 2.$

Given $ABC$ is an equilateral triangle of side length $1$ unit and $P$ be any arbitrary point on the circumcircle of triangle $ABC,$ then $|\vec{PA}|^2+|\vec{PB}|^2+|\vec{PC}|^2$ is equal to

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