If $D(2,1,0), E(2,0,0)$ and $F(0,1,0)$ are mid-points of the sides $B C, C A$ and $A B$ of $\triangle A B C$, respectively. Then, the centroid of $\triangle A B C$ is

• A

  $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$
• Correct answer
  B

  $\left(\frac{4}{3}, \frac{2}{3}, 0\right)$
• C

  $\left(-\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$
• D

  $\left(\frac{2}{3}, \frac{1}{3}, \frac{1}{3}\right)$

Solution

Let $\quad A \equiv\left(x_1, y_1, z_1\right), \quad B \equiv\left(x_2, y_2, z_2\right) \quad$ and $C \equiv\left(x_3, y_3, z_3\right)$
Since, $F$ is the mid-point of $A B$. $$ \left.\therefore \quad \begin{array}{l} x_1+x_2=0 \\ y_1+y_2=2 \\ z_1+z_2=0 \end{array}\right\} $$ Since, $D$ is the mid-point of $B C$. $$ \left.\therefore \quad \begin{array}{l} x_2+x_3=4 \\ y_2+y_3=2 \\ z_2+z_3=0 \end{array}\right\} $$ and $E$ is the mid-point of $A C$ $$ \left.\therefore \quad \begin{array}{l} x_3+x_1=4 \\ y_3+y_1=0 \\ z_3+z_1=0 \end{array}\right\} $$ So, $$ \left.\begin{array}{l} x_1+x_2+x_3=4 \\ y_1+y_2+y_3=2 \\ z_1+z_2+z_3=0 \end{array}\right\} $$ $$ \begin{array}{llll} \therefore & x_3=4, & y_3=0, & z_3=0 \\ & x_1=0, & y_1=0, & z_1=0 \\ \text { and } & x_2=0, & y_2=2, & z_2=0 \end{array} $$ $\therefore$ Centroid of $\triangle A B C$ $$ \begin{aligned} & =\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right) \\ & =\left(\frac{4}{3}, \frac{2}{3}, 0\right) \end{aligned} $$

Hence, the correct answer is option: 'B'.