If the plane meets the coordinate axes in $A, B, C$ such that the centroid of $△ A B C$ is $(p, q, r)$ then the equation of the plane is:

\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 1

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\frac{x}{2 p} + \frac{y}{2 q} + \frac{z}{2 r} = 1

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\frac{x}{3 p} + \frac{y}{3 q} + \frac{z}{3 r} = 1

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\frac{3 x}{p} + \frac{3 y}{q} + \frac{3 z}{r} = 1

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Solution

The correct option is C $\frac{x}{3 p} + \frac{y}{3 q} + \frac{z}{3 r} = 1$
Let the coordinates points $A, B, C$ are $() a, 0, 0), (0, b, 0), (0, 0, c)$ respectively.
then,
Coordinates of centroid of $Δ A B C$

$= (\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$

$(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}) = (p, q, r)$

$a = 3 p, b = 3 q, c = 3 r$
therefore,
equation of the plane is

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

$\frac{x}{3 p} + \frac{y}{3 q} + \frac{z}{3 r} = 1$

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A, B, C

△ A B C

(p, q, r) .

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