Find the equation of a plane which meets the axes at A, B and C, given that the centroid of the triangle ABC is the point (α, β, γ).

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Solution

$Let a, b and c be the intercepts of the given plane on the coordinate axes. Then the plane meets the coordinate axes at A (a, 0, 0), B (0, b, 0) and C (0, 0, c) Given that the centroid of the triangle = (α, β, γ) ⇒ (\frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}, \frac{0 + 0 + c}{3}) = (α, β, γ) ⇒ (\frac{a}{3}, \frac{b}{3}, \frac{c}{3}) = (α, β, γ) ⇒ \frac{a}{3} = α, \frac{b}{3} = β, \frac{c}{3} = γ \Rightarrow a = 3 α, b = 3 β, c = 3 γ . . . (1) The e quation of the plane whose intercepts on the coordinate axes are a, b and c are \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \Rightarrow \frac{x}{3 α} + \frac{y}{3 β} + \frac{z}{3 γ} = 1 [From (1)] \Rightarrow \frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 3$

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