The points A(2, 9, 12), B(1, 8, 8), C(–2, 11, 8) and $D\left(-1,12,12\right)$ form

The given vertices are $A\left(2,9,12\right)$, $B\left(1,8,8\right)$ $C\left(-2,11,8\right)$ and $D\left(-1,12,12\right)$

Observing the options, let us find the lengths of adjacent sides and the lengths of diagonals

Adjacent sides are $AB,AD$

$AB=\sqrt{{\left(1-2\right)}^2+{\left(8-9\right)}^2+{\left(8-12\right)}^2}=\sqrt{1+1+16}=\sqrt{18}$

$AD=\sqrt{{\left(-1-2\right)}^2+{\left(12-9\right)}^2+{\left(12-12\right)}^2}=\sqrt{9+9+0}=\sqrt{18}$

Adjacent sides are equal, so the figure formed by the given vertices is either square or Rhombus ( as in the options)

Consier the lengths of the diagonals 

$AC=\sqrt{{\left(-2-2\right)}^2+{\left(11-9\right)}^2+{\left(8-12\right)}^2}=\sqrt{16+4+16}=6$

$BD=\sqrt{{\left(-1-1\right)}^2+{\left(12-8\right)}^2+{\left(12-8\right)}^2}=\sqrt{4+15+16}=6$

DIagonals are equal  then the figure formed by the given vertices is either square or rectangle

Therefor,e the figure formed by the given vertices is Square.