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(B) Let the plane meet the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.The centroid of $\Delta ABC$ is given by the formula $\left(\

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$x/3 + y/6 + z/9 = 1$

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(B) Let the plane meet the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.The centroid of $\Delta ABC$ is given by the formula $\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right) = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$.Given that the centroid is $(1, 2, 3)$,we have:$\frac{a}{3} = 1 \Rightarrow a = 3$$\frac{b}{3} = 2 \Rightarrow b = 6$$\frac{c}{3} = 3 \Rightarrow c = 9$The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Substituting the values of $a, b$,and $c$,we get:$\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$.

$A$ plane meets the coordinate axes at points $A, B$,and $C$ such that the centroid of $\Delta ABC$ is $(1, 2, 3)$. The equation of the plane is

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