If the plane 56x + 4y + 9z = 2016 meets the c | vedclass

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(C) The given equation of the plane is $56x + 4y + 9z = 2016$. Dividing by $2016$,we get the intercept form: $\frac{x}{36} + \frac{y}{504} + \frac{z}{

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$\left(12, 168, \frac{224}{3}\right)$

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(C) The given equation of the plane is $56x + 4y + 9z = 2016$. Dividing by $2016$,we get the intercept form: $\frac{x}{36} + \frac{y}{504} + \frac{z}{224} = 1$. The plane meets the coordinate axes at $A(36, 0, 0)$,$B(0, 504, 0)$,and $C(0, 0, 224)$. The centroid $G$ of $	riangle ABC$ is given by the formula $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}ight)$. $G = \left(\frac{36+0+0}{3}, \frac{0+504+0}{3}, \frac{0+0+224}{3}ight) = \left(12, 168, \frac{224}{3}ight)$.

If the plane $56x + 4y + 9z = 2016$ meets the coordinate axes at points $A, B$,and $C$,then the centroid of the $\triangle ABC$ is

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