Given that A + B + C = 0. Out of three vectors two are equal in magnitude and the magnitude of the third vector is $\sqrt{2}$ times that of either of the two having equal magnitude. Then the angles between vectors are given by

Let $\left|\mathrm{A}\right|=\left|\mathrm{B}\right|=\mathrm{a}$
then $\left|\mathrm{C}\right|=\sqrt{2}\mathrm{a}$
Given that A+B+C = 0
or A+B = -C
Taking self Product, we have
$\begin{array}{l}(\mathrm{A}+\mathrm{B})\cdot (\mathrm{A}+\mathrm{B})=(-\mathrm{C})\cdot (-\mathrm{C})\\ {\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{A}\cdot \mathrm{B}={\mathrm{C}}^2\\ {\mathrm{a}}^2+{\mathrm{a}}^2+2{\mathrm{a}}^2\cos \mathrm{\theta }=2{\mathrm{a}}^2\\ \cos \mathrm{\theta }=0 \mathrm{or} \mathrm{\theta }={90}^{\circ }\end{array}$
So, angle between A and B - 90"
Again B+C = -A
$\begin{array}{l}\mathrm{or} (\mathrm{B}+\mathrm{C})\cdot (\mathrm{B}+\mathrm{C})=(-\mathrm{A})\cdot (-\mathrm{A})\\ \mathrm{or} {\mathrm{B}}^2+{\mathrm{C}}^2+2\mathrm{B}\cdot \mathrm{C}={\mathrm{A}}^2\\ {\mathrm{a}}^2+2{\mathrm{a}}^2+2\sqrt{2}{\mathrm{a}}^2\cos \mathrm{\varphi }={\mathrm{a}}^2\\ \cos \mathrm{\varphi }=-1/\sqrt{2} \mathrm{or} \mathrm{\varphi }={135}^{\circ }\end{array}$
So, angle between B and C is 135°'
similarly, angle between c and A is 135°.