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Question
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).
Solution
\[\text{ Let a,b and c be the intercepts of the given plane on the coordinate axes.} \]
\[\text{ Then the plane meets the coordinate axes at } \]
\[A \left( a, 0, 0 \right), B \left( 0, b, 0 \right) \text{ and } C \left( 0, 0, c \right)\]
\[\text{ Given that the centroid of the triangle } =\left( \alpha, \beta, \gamma \right)\]
\[\Rightarrow\left( \frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}, \frac{0 + 0 + c}{3} \right)=\left( \alpha, \beta, \gamma \right)\]
\[\Rightarrow\left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right)=\left( \alpha, \beta, \gamma \right)\]
\[\Rightarrow\frac{a}{3}= \alpha,\frac{b}{3}= \beta,\frac{c}{3}= \gamma\]
\[ \Rightarrow a = 3\alpha, b = 3\beta, c = 3\gamma . . . \left( 1 \right)\]
\[\text{ The equation of the plane whose intercepts on the coordinate axes are a ,b and c are } \]
\[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\]
\[ \Rightarrow \frac{x}{3\alpha} + \frac{y}{3\beta} + \frac{z}{3\gamma} = 1 [\text{ From } (1)]\]
\[ \Rightarrow \frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3\]
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