Show that the triangle with vertices \((5, 4, −1)\), \((3, 6, −1)\), and \((3, 4, 1)\) is equilateral.

Given vertices of the triangle are \((5, 4, −1)\), \((3, 6, −1)\), and \((3, 4, 1)\).

Let A= \((5, 4, −1)\), B= \((3, 6, −1)\), and C=\((3, 4, 1)\)

Distance AB:

A= (5, 4, −1), B= (3, 6, −1)

AB = \(\sqrt {{{(3 - 5)}^2} + {{\left( {6 - 4} \right)}^2} + {{\left( { - 1 + 1} \right)}^2}} \)

=\(\sqrt {4 + 4 + 0} \)

= \(\sqrt 8 \) units

Distance BC:

B= (3, 6, −1) ,C=(3, 4, 1)

BC =\(\sqrt {{{(3 - 3)}^2} + {{\left( {4 - 6} \right)}^2} + {{\left( { 1 + 1} \right)}^2}} \)

= \(\sqrt {0 + 4 + 4} \)

= \(\sqrt 8 \) units

Distance AC:

A= (5, 4, −1) ,C=(3, 4, 1)

AC = \(\sqrt {{{(3 - 5)}^2} + {{\left( {4 - 4} \right)}^2} + {{\left( { 1 + 1} \right)}^2}} \)

=\(\sqrt {4 + 0 + 4} \)

= \(\sqrt 8 \) units

Here the sides of triangle ABC are equal in length.

Therefore, the given triangle is an equilateral triangle.