A plane meet the co-ordinates axes in $A, B, C$ such that the centroid of triangle $A B C$ is the point $α, β, γ .$ If the equation of the plane be $\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = k$ then, $k = ?$

1

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3

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2

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α^{2} + β^{2} + γ^{2}

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Solution

The correct option is B $3$
Equation of plane ABC in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 . . . . . (i)$
So $A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c)$
Also centroid of $△ A B C i s (\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$
But co-ordinates of centroid is $(α, β, γ)$
$∴ α = \frac{a}{3}$
$∴ a = 3 α, b = 3 β, c = 3 γ$
Putting in (i) we get
$\frac{x}{3 α} + \frac{y}{3 β} + \frac{y}{3 γ} = 1$
$\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 3$

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