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

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane is at a unit distance from the origin $(0, 0, 0)$,the perpendicular distance $d$ is given by $d = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 1$. Squaring both sides,we get $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = 1$ ... $(i)$. The coordinates of the points $P, Q,$ and $R$ are $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ respectively. The centroid $(x, y, z)$ of $\Delta PQR$ is given by $x = \frac{a+0+0}{3} = \frac{a}{3}$,$y = \frac{0+b+0}{3} = \frac{b}{3}$,and $z = \frac{0+0+c}{3} = \frac{c}{3}$. Thus,$a = 3x, b = 3y, c = 3z$. Substituting these into equation $(i)$,we get $\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = 1$. This simplifies to $\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = 1$,which implies $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 9$. Comparing this with the given locus $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = k$,we find $k = 9$.

$A$ plane is at a unit distance from the origin. It cuts the coordinate axes at $P, Q,$ and $R$ respectively. If the locus of the centroid of the $\Delta PQR$ is $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = k$,then $k =$

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