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(D) Let the position vectors of the points $A, B,$ and $C$ be: $\vec{OA} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ $\vec{OB} = \beta \hat{i}

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vertices of an equilateral triangle

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(D) Let the position vectors of the points $A, B,$ and $C$ be: $\vec{OA} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ $\vec{OB} = \beta \hat{i} + \gamma \hat{j} + \alpha \hat{k}$ $\vec{OC} = \gamma \hat{i} + \alpha \hat{j} + \beta \hat{k}$ The vectors representing the sides are: $\vec{AB} = \vec{OB} - \vec{OA} = (\beta - \alpha) \hat{i} + (\gamma - \beta) \hat{j} + (\alpha - \gamma) \hat{k}$ $\vec{BC} = \vec{OC} - \vec{OB} = (\gamma - \beta) \hat{i} + (\alpha - \gamma) \hat{j} + (\beta - \alpha) \hat{k}$ $\vec{CA} = \vec{OA} - \vec{OC} = (\alpha - \gamma) \hat{i} + (\beta - \alpha) \hat{j} + (\gamma - \beta) \hat{k}$ Now,calculate the magnitudes of these sides: $|\vec{AB}| = \sqrt{(\beta - \alpha)^2 + (\gamma - \beta)^2 + (\alpha - \gamma)^2}$ $|\vec{BC}| = \sqrt{(\gamma - \beta)^2 + (\alpha - \gamma)^2 + (\beta - \alpha)^2}$ $|\vec{CA}| = \sqrt{(\alpha - \gamma)^2 + (\beta - \alpha)^2 + (\gamma - \beta)^2}$ Since $|\vec{AB}| = |\vec{BC}| = |\vec{CA}|$,the points form an equilateral triangle. Since $\alpha, \beta, \gamma$ are distinct,the magnitude is non-zero. Thus,the points are vertices of an equilateral triangle.

If $\alpha, \beta, \gamma$ are distinct real numbers and $\alpha+\beta+\gamma \neq 0$,then the points with position vectors $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}, \beta \hat{i}+\gamma \hat{j}+\alpha \hat{k}$ and $\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k}$ are

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