The coplanar points (3, 2, 5), (2, 1, 1), (–1, 4, 1), (0, 5, 5) forms a

The given points are $A\left(3,2,5\right),B\left(2,1,1\right),C\left(-1,4,1\right),D\left(0,5,5\right)$

Observing the options, let us find the lengths of adjacent sides $AB,AD$ and diagonalas $AC,BD$

Hence, $AB=\sqrt{{\left(2-3\right)}^2+{\left(1-2\right)}^2+{\left(1-5\right)}^2}=\sqrt{1+1+16}=\sqrt{18}$

$AD=\sqrt{{\left(0-3\right)}^2+{\left(5-2\right)}^2+{\left(5-5\right)}^2}=\sqrt{9+9+0}=\sqrt{18}$

Since the adjacent sides are equal, so that the figure formed by the given vertices are either square or Rhombus

Let us find the lengths of the diagonals

$AC=\sqrt{{\left(-1-3\right)}^2+{\left(4-2\right)}^2+{\left(1-5\right)}^2}=\sqrt{16+4+16}=6$

$BD=\sqrt{{\left(0-2\right)}^2+{\left(5-1\right)}^2+{\left(5-1\right)}^2}=\sqrt{16+4+16}=6$

Diagonals are also equal

Hence, the figure formed by the given four points is square.