$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the $\Delta ABC$ is the point $(\alpha, \beta, \gamma)$. Show that the equation of the plane is $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$.

(A) Let the equation of the plane in intercept form be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
The coordinates of the points $A, B, C$ where the plane meets the axes are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.
The centroid of $\Delta ABC$ is given by the formula $\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$.
Substituting the coordinates of $A, B, C$,the centroid is $\left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right)$.
We are given that the centroid is $(\alpha, \beta, \gamma)$.
Equating the coordinates,we get $\alpha = \frac{a}{3}$,$\beta = \frac{b}{3}$,and $\gamma = \frac{c}{3}$.
This implies $a = 3\alpha$,$b = 3\beta$,and $c = 3\gamma$.
Substituting these values into the intercept form equation $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,we get $\frac{x}{3\alpha} + \frac{y}{3\beta} + \frac{z}{3\gamma} = 1$.
Multiplying both sides by $3$,we obtain the required equation: $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$.