If $\hat{a},\hat{b}$ and $\hat{c}$ are unit vectors, then $|\hat{a}-\hat{b}{|}^2+|\hat{b}-\hat{c}{|}^2+|\hat{c}-\hat{a}{|}^2$ does not exceed

$\hat{a},\hat{b}$ and $\hat{c}$ are unit vectors, Now

$x=|\hat{a}-\hat{b}{|}^2+|\hat{b}-\hat{c}{|}^2+|\hat{c}-\hat{a}{|}^2$

$=\frac{1}{2}(\hat{a}\cdot \hat{a}+\hat{b}\cdot \hat{b}+\hat{c}\cdot \hat{c})-2\hat{a}\cdot \hat{b}-2\hat{b}\cdot \hat{c}-2\hat{c}\cdot \hat{a}$$=6-2(\hat{a}.\hat{b}+\hat{b}.\hat{c}+\hat{c}.\hat{a})$

Also $|\hat{a}+\hat{b}+\hat{c}|\ge 0$

Or  $\hat{a}\cdot \hat{a}+\hat{b}\cdot \hat{b}+\hat{c}.\hat{c}+2(\hat{a}\cdot \hat{b}+\hat{b}\cdot \hat{c}+\hat{c}.\hat{a})\ge 0$

Or  $3+2(\hat{a}\cdot \hat{b}+\hat{b}\cdot \hat{c}+\hat{c}\cdot \hat{a})\ge 0$

Or $-2(\hat{a}\cdot \hat{b}+\hat{b}\cdot \hat{c}+\hat{c}.\hat{a})\le 3$

Or $6-2(\hat{a}\cdot \hat{b}+\hat{b}\cdot \hat{c}+\hat{c}.\hat{a})\le 9$

From (i) and (ii),  $x\le 0$

Therefore, x does not exceed 9 .