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(B) Let the points be $P$,$Q$,and $R$ with position vectors $\vec{p} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,$\vec{q} = \beta \hat{i} + \ga

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an equilateral triangle

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(B) Let the points be $P$,$Q$,and $R$ with position vectors $\vec{p} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,$\vec{q} = \beta \hat{i} + \gamma \hat{j} + \alpha \hat{k}$,and $\vec{r} = \gamma \hat{i} + \alpha \hat{j} + \beta \hat{k}$.The distance $PQ = |\vec{q} - \vec{p}| = \sqrt{(\beta - \alpha)^2 + (\gamma - \beta)^2 + (\alpha - \gamma)^2}$.The distance $QR = |\vec{r} - \vec{q}| = \sqrt{(\gamma - \beta)^2 + (\alpha - \gamma)^2 + (\beta - \alpha)^2}$.The distance $RP = |\vec{p} - \vec{r}| = \sqrt{(\alpha - \gamma)^2 + (\beta - \alpha)^2 + (\gamma - \beta)^2}$.Since $PQ = QR = RP$,the points form an equilateral triangle.

Let $\alpha, \beta, \gamma$ be distinct real numbers. The points with position vectors $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,$\beta \hat{i} + \gamma \hat{j} + \alpha \hat{k}$,and $\gamma \hat{i} + \alpha \hat{j} + \beta \hat{k}$ form:

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