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(D) Let the equation of the variable plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.This plane meets the coordinate axes at $A(a, 0, 0)$,$B(0,

Vedclass · Accepted Answer

$x^{-2} + y^{-2} + z^{-2} = 9k^{-2}$

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(D) Let the equation of the variable plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.This plane meets the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.Let $(\alpha, \beta, \gamma)$ be the coordinates of the centroid of $\Delta ABC$. Then,$\alpha = \frac{a}{3}, \beta = \frac{b}{3}, \gamma = \frac{c}{3} \implies a = 3\alpha, b = 3\beta, c = 3\gamma \dots (i)$The distance of the plane from the origin is $k$.$	herefore \left| \frac{\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} ight| = k$$\implies \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{k^2}$Substituting values from $(i)$:$\frac{1}{(3\alpha)^2} + \frac{1}{(3\beta)^2} + \frac{1}{(3\gamma)^2} = \frac{1}{k^2}$$\frac{1}{9\alpha^2} + \frac{1}{9\beta^2} + \frac{1}{9\gamma^2} = \frac{1}{k^2}$$\alpha^{-2} + \beta^{-2} + \gamma^{-2} = 9k^{-2}$Thus,the locus of $(\alpha, \beta, \gamma)$ is $x^{-2} + y^{-2} + z^{-2} = 9k^{-2}$.

$A$ variable plane is at a distance $k$ from the origin and meets the coordinate axes at $A, B, C$. The locus of the centroid of $\Delta ABC$ is . . . . . .

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