If the plane 7x + 11y + 13z = 3003 meets the | vedclass

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(A) The given equation of the plane is $7x + 11y + 13z = 3003$. Dividing the entire equation by $3003$, we get the intercept form of the plane: $\frac

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$(143, 91, 77)$

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(A) The given equation of the plane is $7x + 11y + 13z = 3003$. Dividing the entire equation by $3003$, we get the intercept form of the plane: $\frac{7x}{3003} + \frac{11y}{3003} + \frac{13z}{3003} = 1$ $\Rightarrow \frac{x}{429} + \frac{y}{273} + \frac{z}{231} = 1$. The plane meets the coordinate axes at points $A, B,$ and $C$. Setting $y=0, z=0$ gives $x=429$, so $A = (429, 0, 0)$. Setting $x=0, z=0$ gives $y=273$, so $B = (0, 273, 0)$. Setting $x=0, y=0$ gives $z=231$, so $C = (0, 0, 231)$. The centroid of $	riangle ABC$ is given by the formula $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}ight)$. Centroid $= \left(\frac{429+0+0}{3}, \frac{0+273+0}{3}, \frac{0+0+231}{3}ight)$ $= (143, 91, 77)$.

If the plane $7x + 11y + 13z = 3003$ meets the coordinate axes at $A, B, C$, then the centroid of the $\triangle ABC$ is

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