Points A(0,1,2) , B(2,0,1) and C(3,1,0) are three…

Points $A(0,1,2)$, $B(2,0,1)$ and $C(3,1,0)$ are three consecutive vertices of a parallelogram. Find point $O'$ — the intersection of the diagonals — and the centroid of triangle $ABC$.

Step 1: Fourth vertex $D$

\[ \overrightarrow{BC} = (1,1,-1), \qquad D = A + \overrightarrow{BC} = (0,1,2)+(1,1,-1) = (1,2,1) \]

Check: $\overrightarrow{AB} = (2,-1,-1) = \overrightarrow{DC}$ $\checkmark$

Step 2: Intersection of the diagonals

The diagonals bisect each other, so $O'$ is the midpoint of both: \[\text{mid}(AC) = \left(\frac{0+3}{2},\,\frac{1+1}{2},\,\frac{2+0}{2}\right) = \left(\frac{3}{2},1,1\right), \qquad \text{mid}(BD) = \left(\frac{3}{2},1,1\right) \;\checkmark\]

\[O'\left(\frac{3}{2},\;1,\;1\right)\]

(Since $O'$ is the midpoint of $AC$, it can be found without $D$ at all.)

Step 3: Centroid of triangle $ABC$

\[ G = \frac{A+B+C}{3} = \left(\frac{0+2+3}{3},\,\frac{1+0+1}{3},\,\frac{2+1+0}{3}\right) \]

\[G\left(\frac{5}{3},\;\frac{2}{3},\;1\right)\]

Check via the median

The midpoint of $BC$ is $A_1\left(\frac52,\frac12,\frac12\right)$, and $G$ divides $AA_1$ in the ratio $2:1$: \[G = A + \frac{2}{3}\overrightarrow{AA_1} = (0,1,2) + \frac{2}{3}\left(\frac{5}{2},-\frac{1}{2},-\frac{3}{2}\right) = \left(\frac{5}{3},\frac{2}{3},1\right) \;\checkmark\]