The perimeter of triangle with vertices at (1,0,0).(0,1,0) and (0,0,1) is:-
a.)3 b.) 2 c.) $2\sqrt 2 $ d.) $3\sqrt 2 $

By using distance formula, we would find the sides, AB, BC, and AC;
So, AB $ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2} + {{\left( {{z_1} - {z_2}} \right)}^2}} $
     $ = \sqrt {{{\left( {0 - 1} \right)}^2} + {{\left( {1 - 0} \right)}^2} + {{\left( {0 - 0} \right)}^2}} $
     $ = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2} + {0^2}} $
     $ = \sqrt {1 + 1 + 0} = \sqrt 2 $
Similarly, BC $ = \sqrt {{{\left( {{x_3} - {x_2}} \right)}^2} - {{\left( {{y_3} - {y_2}} \right)}^2} + {{\left( {{z_3} - {z_2}} \right)}^2}} $
          $ = \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( {0 + 1} \right)}^2} + {{\left( {1 - 0} \right)}^2}} $
          $ = \sqrt {{0^2} + {{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2}} $
          =$\sqrt 2 $
Similarly, AC = $ = \sqrt {{{\left( {{x_3} - {x_1}} \right)}^2} + {{\left( {{y_3} - {y_1}} \right)}^2} + {{\left( {{z_3} - {z_1}} \right)}^2}} $
          $ = \sqrt {{{\left( {0 - 1} \right)}^2} + {{\left( {0 - 0} \right)}^2} + {{\left( {1 - 0} \right)}^2}} $
          $ = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 0 \right)}^2} + {{\left( 1 \right)}^2}} $
          $\sqrt {1 + 0 + 1} \;\; = \,\sqrt 2 $
Hence, perimeter of triangle is given by:-
     AB + BC + AC
     $ = \sqrt 2 + \sqrt 2 + \sqrt 2 $
     $3\sqrt 2 $

Note:- The distance formula for the three coordinate plane with points ${\text{A}}\left( {{x_1},{y_1},{z_1}} \right)$ and B $\left( {{x_2},{y_2},{z_2}} \right)$ are:-
AB $ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $