Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

(D) Given equation of plane can be rewritten as $\frac{\mathrm{x}}{\left(\frac{1}{2}\right)}+\frac{\mathrm{y}}{\left(\frac{1}{3}\right)}+\frac{\mathrm{z}}{\left(\frac{1}{5}\right)}=1$ i.e. intercepts on $\mathrm{X},\mathrm{Y},\mathrm{Z}$ axis are $\frac{1}{2},\frac{1}{3},\frac{1}{5}$ respectively. $\mathrm{A}=\left(\frac{1}{2},0,0\right),\mathrm{B}=\left(0,\frac{1}{3},0\right),\mathrm{C}=\left(0,0,\frac{1}{5}\right)$ Thus centroid of $△\mathrm{ABC}=\left(\frac{\frac{1}{2}+0+0}{3},\frac{0+\frac{1}{3}+0}{3},\frac{0+0+\frac{1}{5}}{3}\right)=\left(\frac{1}{6},\frac{1}{9},\frac{1}{15}\right)$