Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Let the equation of the plane is $\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$ Then, $\mathrm{A}(\alpha ,0,0),\mathrm{B}(0,\beta ,0)$ and $\mathrm{C}(0,0,\gamma )$ are the points on the coordinate axes, The centroid of the triangle is $(1,2,4)$. $\begin{array}{l}\therefore \frac{\alpha }{3}=1\Rightarrow \alpha =3\\ \frac{\beta }{3}=2\Rightarrow \beta =6\end{array}$ and $\frac{\gamma }{3}=4\Rightarrow \gamma =12$ $\therefore$ The equation of the plane is $\begin{array}{rl}& \frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1\\ \Rightarrow & 4x+2y+z=12\end{array}$