If the plane \(2 x+3 y+5 z=1\) intersects the co-ordinate axes at the points \(A, B, C\), then the centroid of \(\triangle \mathrm{ABC}\) is

(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 \(\triangle \mathrm{ABC}=\left(\frac{\frac{1}{2}+0+0}{3}, \frac{0+\frac{1}{3}+0}{3}, \frac{0+0+\frac{1}{5}}{3}\right)=\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{15}\right)\)