# Let a , b anc c be distinct and real numbers . The points with position vectors ai+bj+ck , bi+cj+ak and ci+aj+bk Then

Solution:   Let the given  position vectors  be of points

A , B and C respectively , then

$\\ \\ \Rightarrow \hspace{1cm}\left | \vec{AB} \right |=\sqrt{(b-a)^{2}+(c-b)^{2}+(a-c)^{2}}\\ \\ \Rightarrow \hspace{1cm}\left | \vec{BC} \right |=\sqrt{(c-b)^{2}+(a-c)^{2}+(a-b)^{2}}\\ \\ \Rightarrow \hspace{1cm}\left | \vec{CA} \right |=\sqrt{(a-c)^{2}+(b-a)^{2}+(c-b)^{2}}\\ \\ \therefore \hspace{1cm}\left | \vec{AB} \right |=\left | \vec{BC} \right |=\left | \vec{CA} \right |$

Hence ,    $\bigtriangleup ABC$ is an  equilateral triangle

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