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(D) Let the magnitudes of the vectors be $|A| = x$,$|B| = x$,and $|C| = \sqrt{2}x$.Since $\overrightarrow A + \overrightarrow B + \overrightarrow C =

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$90^{\circ}, 135^{\circ}, 135^{\circ}$

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(D) Let the magnitudes of the vectors be $|A| = x$,$|B| = x$,and $|C| = \sqrt{2}x$.Since $\overrightarrow A + \overrightarrow B + \overrightarrow C = 0$,these three vectors form a closed triangle.According to the law of cosines for a triangle with sides $x, x, \sqrt{2}x$,we observe that $x^2 + x^2 = (\sqrt{2}x)^2$,which satisfies the Pythagorean theorem.Thus,the vectors form a right-angled isosceles triangle.To find the angles between the vectors,we place them head-to-tail.The angle between $\overrightarrow A$ and $\overrightarrow B$ is $90^{\circ}$ (as they are the legs of the right triangle).The angle between $\overrightarrow B$ and $\overrightarrow C$ is $180^{\circ} - 45^{\circ} = 135^{\circ}$.The angle between $\overrightarrow A$ and $\overrightarrow C$ is $180^{\circ} - 45^{\circ} = 135^{\circ}$.Therefore,the angles are $90^{\circ}, 135^{\circ}, 135^{\circ}$.

Given that $\overrightarrow A + \overrightarrow B + \overrightarrow C = 0$. 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 the vectors are given by:

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