Solution: Given that a plane meets coordinate axes at points $A,B$ and $C$. So we assume that the coordinates of $A=(a,0,0), B= (0,b,0)$ and $C=(0,0,c)$. We need to find the equation of the plane. Recall that the equation of the plane which intersects axes at $(a,0,0),(0,b,0)$ and $(0,0,c)$ is given by \[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1. \] We have the following picture:

Recall that the centroid of the triangle with vertices $(x_i,y_i,z_i),~ i = 1,2,3$ is given by \[ \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right). \] Therefore, the centroid of the given triangle will be \[ \left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) = \left( \frac{a}{3},\frac{b}{3}, \frac{c}{3} \right) . \] From the given data in the problem, we have \begin{align*} \left( \frac{a}{3},\frac{b}{3}, \frac{c}{3} \right) = \left( 1,\frac{1}{2},-\frac{1}{3} \right) \implies a=3,~b=\frac{3}{2} \text{ and } c = -1. \end{align*} Therefore, the equation of the plane will be \[ \frac{x}{a}+ \frac{y}{b} + \frac{z}{c} = 1 \implies \frac{x}{3} + \frac{2y}{3} - z = 1. \]