Let α, β, γ be distinct real numbers. The points with position...

$\alpha, \beta$ and $\gamma$ be distinct real numbers $$\begin{aligned} & \vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k} \\ & \vec{b}=\beta \hat{i}+\gamma \hat{j}+\alpha \hat{k} \\ & \vec{c}=\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k} \end{aligned}$$ Let $\alpha=1, \beta=2$ and $\alpha=3$ then, $\quad \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$$\begin{aligned} & \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \end{aligned}$$$Now,$|\overrightarrow{\mathrm{ab}}|=\sqrt{(2-1)^2}+(3-2)^2+(1-3)^2=\sqrt{6}$Similarly$|\overrightarrow{\mathrm{bc}}|=|\overrightarrow{\mathrm{ac}}|=\sqrt{6}$ Hence, point for equilateral triangle.