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(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane is at a constant distance $h$ from the origin $(0,

Vedclass · Accepted Answer

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{9}{h^2}$

Vedclass · Answer

(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane is at a constant distance $h$ from the origin $(0, 0, 0)$,we have $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = h$,which implies $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{h^2}$. The coordinates of $A, B, C$ are $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ respectively. Let $(x, y, z)$ be the centroid of $	riangle ABC$. Then $x = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3}$. This gives $a = 3x, b = 3y, c = 3z$. Substituting these into the plane distance equation: $\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = \frac{1}{h^2}$ $\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = \frac{1}{h^2}$ $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{9}{h^2}$.

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

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