Let P is any arbitrary point on the circumcircle of a given equilateral triangle of side length ' $\ell$ ' units then, $|\vec{P A}|^2+|\vec{P B}|^2+|\vec{P C}|^2$ is always equal to:

$2\ell^2$

Let P.V. of P, A, B and C are $\vec{p}, \vec{a}, \vec{b}$ and $\vec{c}$ respectively and $O(\vec{0})$ be the circumcentre of the equilateral triangle ABC.

$\Rightarrow|\vec{p}|=|\vec{b}|=|\vec{a}|=|\vec{c}|=\frac{\ell}{\sqrt{3}}$

Now, $|\vec{P A}|^2=|\vec{a}-\vec{p}|^2=|\vec{a}|^2+|\vec{p}|^2-2 \vec{p} . \vec{a}$

Similarly, $|\vec{P B}|^2=|\vec{b}|^2+|\vec{p}|^2-2 \vec{p} . \vec{b}$

and $|\vec{P C}|^2=|\vec{c}|^2+|\vec{p}|^2-2 \vec{p} . \vec{c}$

$\Rightarrow \sum|\vec{P A}|^2=6 . \frac{\ell^2}{3}-2 \vec{p} . (\vec{a}+b+\vec{c})$

$=2 \ell^2$,  as  $\frac{\vec{a}+\vec{b}+\vec{c}}{3}=0$

Hence (1) is correct answer.