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Centroid of a Triangle

The plane ax ...

Question

The plane ax + by + cz = 1 meets the coordinate axes in A, B and C. The centroid of $△ A B C$ is

(3 a, 3 b, 3 c)

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

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

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(\frac{1}{3 a}, \frac{1}{3 b}, \frac{1}{3 c})

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Solution

The correct option is D $(\frac{1}{3 a}, \frac{1}{3 b}, \frac{1}{3 c})$
Coordinates of A,B,C are $(\frac{1}{a}, 0), (\frac{1}{b}, 0), (\frac{1}{c}, 0)$ respectively
centriod of $Δ A B C = (\frac{1}{3 a}, \frac{1}{3 b}, \frac{1}{3 c})$

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