$ for point $A$

$ \mathbf{r}_2 = 2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} - \hat{\mathbf{k}} $ for point $B$

$ \mathbf{r}_3 = -3 \hat{\mathbf{i}} - \hat{\mathbf{j}} - 2 \hat{\mathbf{k}} $ for point $C$

We can calculate the distances between these points:

Distance between $A$ and $B$:

$ \begin{aligned} |\mathbf{A B}| &= \left|\mathbf{r}_2 - \mathbf{r}_1\right| \\ &= |(2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} - \hat{\mathbf{k}}) - (\hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}})| \\ &= |\hat{\mathbf{i}} + 5 \hat{\mathbf{j}} - 4 \hat{\mathbf{k}}| \\ &= \sqrt{1^2 + 5^2 + 4^2} \\ &= \sqrt{1 + 25 + 16} \\ &= \sqrt{42} \end{aligned} $

Distance between $B$ and $C$:

$ \begin{aligned} |\mathbf{B C}| &= \left|\mathbf{r}_3 - \mathbf{r}_2\right| \\ &= |(-3 \hat{\mathbf{i}} - \hat{\mathbf{j}} - 2 \hat{\mathbf{k}}) - (2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} - \hat{\mathbf{k}})| \\ &= |-5 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} - \hat{\mathbf{k}}| \\ &= \sqrt{5^2 + 4^2 + 1^2} \\ &= \sqrt{25 + 16 + 1} \\ &= \sqrt{42} \end{aligned} $

Distance between $C$ and $A$:

Since

$ \mathbf{C A} = \mathbf{r}_1 - \mathbf{r}_3 = (\hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}) - (-3 \hat{\mathbf{i}} - \hat{\mathbf{j}} - 2 \hat{\mathbf{k}}) $

you find

$ |\mathbf{C A}| = \sqrt{42}. $

All three distances are equal,

$ |\mathbf{AB}| = |\mathbf{BC}| = |\mathbf{CA}| = \sqrt{42} $

Thus, points $A$, $B$, and $C$ form an equilateral triangle because all three sides are of equal length.

Vector Algebra

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