Given that $\mathbf{P}+\mathbf{Q}+\mathbf{R}=0$. Two out of the three vectors are equal in magnitude. The magnitude of the third vector is $\sqrt{2}$ times that of the other two. Which of the following can be the angles between these vectors?

P+Q=-R

From cosine rule,

$\cos \alpha =\frac{|P{|}^2+|R{|}^2-|Q{|}^2}{2\left|P\right|\left|\right|R|}$

If $\left|\mathrm{P}\right|=\left|\mathrm{Q}\right|and\left|R\right|=\sqrt{2}\left|P\right|$

Then, $\cos \alpha =\frac{2|P{|}^2}{2|P{|}^2·\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\alpha =45°$ and $\sin \theta$ two angles are equal.

If $\left|\mathrm{R}\right|=\left|\mathrm{Q}\right|and\left|P\right|=\sqrt{2}\left|R\right|$

Remaining angles are $45°,90°$ Then, $\left|P\right|=\left|R\right|and\left|Q\right|=\sqrt{2}\left|P\right|$

$\alpha =90°$ and remaining two angles are $\alpha =45°$ each.