Explanation:
B We know that,
Equation of plane
\(\frac{\mathrm{x}}{\mathrm{x}_1}+\frac{\mathrm{y}}{\mathrm{y}_1}+\frac{\mathrm{z}}{\mathrm{z}_1}=1\)
\(\therefore \quad\)
It meets the coordinate-axes on the points\(\mathrm{A}\left(\mathrm{x}_1, 0,0\right), \mathrm{B}\left(0, \mathrm{y}_1, 0\right)\) and \(\mathrm{C}\left(0,0, \mathrm{z}_1\right)\)
Then the centroid of \(\triangle \mathrm{ABC}\) is-
\(=\left(\frac{\mathrm{x}_1+0+0}{3}, \frac{0+\mathrm{y}_1+\mathrm{y}}{3}, \frac{0+0+\mathrm{z}_1}{3}\right)\)
\(=\left(\frac{\mathrm{x}_1}{3}, \frac{\mathrm{y}_1}{3}, \frac{\mathrm{z}_1}{3}\right)\)
centroid, \((1,2,3)(\because\) Given \()\)
Then,
\(\because \frac{\mathrm{x}_1}{3}=1, \frac{\mathrm{y}_1}{3}=2, \frac{\mathrm{z}_1}{3}=3\)
\(\Rightarrow \mathrm{x}_1=3, \mathrm{y}_1=6, \mathrm{z}_1=9\)
Therefore, the equation (i) of the plane becomes-
\(\frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{6}+\frac{\mathrm{z}}{9}=1\)