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{7x

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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$$\frac{x}{429} + \frac{y}{273} + \frac{z}{231} = 1$The plane meets the coordinate axes at points $A, B,$ and $C$. These points are the intercepts on the $x, y,$ and $z$ axes respectively:$A = (429, 0, 0)$$B = (0, 273, 0)$$C = (0, 0, 231)$The centroid of $	riangle ABC$ with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2),$ and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}ight)$.Centroid $= \left(\frac{429+0+0}{3}, \frac{0+273+0}{3}, \frac{0+0+231}{3}ight)$Centroid $= (143, 91, 77)$.

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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