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