The points A (1, 1, 0), B(0, 1, 1), C(1, 0, 1) and D(2/3, 2/3, 2/3)

The correct option is Coplanar.

To determine if the four given points $A(1,1,0)$, $B(0,1,1)$, $C(1,0,1)$, and $D(\frac{2}{3},\frac{2}{3},\frac{2}{3})$ are coplanar, we can construct three vectors sharing a common initial point (for example, point A) and check if their scalar triple product is equal to zero.

First, let's find the components of the vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$:
For $\overrightarrow{AB}$:
$\overrightarrow{AB}=(0-1)\hat{i}+(1-1)\hat{j}+(1-0)\hat{k}=-1\hat{i}+0\hat{j}+1\hat{k}$
For $\overrightarrow{AC}$:
$\overrightarrow{AC}=(1-1)\hat{i}+(0-1)\hat{j}+(1-0)\hat{k}=0\hat{i}-1\hat{j}+1\hat{k}$
For $\overrightarrow{AD}$:
$\overrightarrow{AD}=(\frac{2}{3}-1)\hat{i}+(\frac{2}{3}-1)\hat{j}+(\frac{2}{3}-0)\hat{k}=-\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}+\frac{2}{3}\hat{k}$

Four points are coplanar if and only if the scalar triple product of the three vectors formed by them is zero, i.e., $\left[\overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD}\right]=0$. We calculate this using the determinant of the matrix containing these vector components:
$det\left(\begin{array}{ccc}-1& 0& 1\\ 0& -1& 1\\ -\frac{1}{3}& -\frac{1}{3}& \frac{2}{3}\end{array}\right)$

Now, expanding the determinant along the first row:
$=-1\left[\right(-1\left)\right(\frac{2}{3})-(1\left)\right(-\frac{1}{3}\left)\right]-0+1\left[\right(0\left)\right(-\frac{1}{3})-(-1\left)\right(-\frac{1}{3}\left)\right]$
Simplify the terms inside the brackets:
$=-1[-\frac{2}{3}+\frac{1}{3}]+1[0-\frac{1}{3}]$
$=-1[-\frac{1}{3}]+1[-\frac{1}{3}]$
$=\frac{1}{3}-\frac{1}{3}=0$

Since the scalar triple product is exactly equal to zero, the vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$ lie in the same plane. Therefore, the given points A, B, C, and D are coplanar.