If a plane meets the axes $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ in $\mathrm{A}, \mathrm{B}, \mathrm{C}$…
If a plane meets the axes $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ in $\mathrm{A}, \mathrm{B}, \mathrm{C}$ respectively such that centroid of $\triangle \mathrm{ABC}$ is $(1,2,3)$, then the equation of the plane is
$x+2 y+3 z=1$
$x+\frac{y}{2}+\frac{z}{3}=3$
$\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$
$\frac{x}{4}+\frac{y}{8}+\frac{z}{12}=1$
Solution
Let $A=(x, 0,0) ; B=(0, y, 0) ; C=(0,0, z)$ Centroid of $\triangle A B C$ is $(1,2,3)$
$\therefore \frac{\mathrm{x}}{3}=1, \frac{\mathrm{y}}{3}=2, \frac{\mathrm{z}}{3}=3 \Rightarrow(\mathrm{x}, \mathrm{y}, \mathrm{z})=(3,6,9)$
Hence equation of required plane is
$\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$