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(B) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ ... $(i)$Since the plane meets the coordinate axes at $A(a, 0, 0)$,

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$4x+2y+z=12$

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(B) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ ... $(i)$Since the plane meets the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$,the centroid of $	riangle ABC$ is given by $\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}ight) = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}ight)$.Given that the centroid is $(1, 2, 4)$,we have:$\frac{a}{3} = 1 \Rightarrow a = 3$$\frac{b}{3} = 2 \Rightarrow b = 6$$\frac{c}{3} = 4 \Rightarrow c = 12$Substituting these values into equation $(i)$,we get:$\frac{x}{3} + \frac{y}{6} + \frac{z}{12} = 1$Multiplying by $12$,we get $4x + 2y + z = 12$.

$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$. Then,the equation of the plane is

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