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

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$x+2 y+3 z=3$

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(B) Let the equation of the plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.Since the plane meets the coordinate axes at $P, Q, R$ respectively, the coordinates of the points $P, Q, R$ are $(a, 0, 0), (0, b, 0), (0, 0, c)$ respectively.The centroid of $	riangle P Q R$ 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 $\left(1, \frac{1}{2}, \frac{1}{3}ight)$, we equate the coordinates:$\frac{a}{3} = 1 \Rightarrow a = 3$$\frac{b}{3} = \frac{1}{2} \Rightarrow b = \frac{3}{2}$$\frac{c}{3} = \frac{1}{3} \Rightarrow c = 1$Substituting these values into the intercept form of the plane equation:$\frac{x}{3} + \frac{y}{3/2} + \frac{z}{1} = 1$$\frac{x}{3} + \frac{2y}{3} + z = 1$Multiplying by $3$, we get $x + 2y + 3z = 3$.

$A$ plane meets the coordinate axes at $P, Q, R$ respectively. If the centroid of $\triangle P Q R$ is $\left(1, \frac{1}{2}, \frac{1}{3}\right)$, then the equation of the plane is

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