The plane frac{x}{a} + frac{y}{b} + frac{z}{c | vedclass

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(D) The equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3$.To find the intersection with the $x$-axis,set $y=0$ and $z=0$: $\frac{

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$(a, b, c)$

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(D) The equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3$.To find the intersection with the $x$-axis,set $y=0$ and $z=0$: $\frac{x}{a} = 3 \implies x = 3a$. So,$A = (3a, 0, 0)$.To find the intersection with the $y$-axis,set $x=0$ and $z=0$: $\frac{y}{b} = 3 \implies y = 3b$. So,$B = (0, 3b, 0)$.To find the intersection with the $z$-axis,set $x=0$ and $y=0$: $\frac{z}{c} = 3 \implies z = 3c$. So,$C = (0, 0, 3c)$.The centroid of a triangle with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2),$ and $(x_3, y_3, z_3)$ is given by $\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$.Substituting the coordinates of $A, B,$ and $C$:Centroid $= \left( \frac{3a+0+0}{3}, \frac{0+3b+0}{3}, \frac{0+0+3c}{3} \right) = (a, b, c)$.

The plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3$ meets the coordinate axes at points $A, B,$ and $C$. The centroid of the triangle $ABC$ is:

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