A variable plane is at a constant distance $p$ from the origin and meets the axes at $A, B, C$. If the locus of the centroid of the tetrahedron $OABC$ is $x^{-2} + y^{-2} + z^{-2} = 2qp^{-2}$ then the value of $\sqrt{\lambda}$ is __________.
Step-by-Step Solution
Key Concept: The centroid formula and distance constraint from origin to plane together determine the relationship between the intercepts and centroid coordinates.
Let the plane equation be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, meeting axes at $A(a,0,0)$, $B(0,b,0)$, $C(0,0,c)$. The distance from origin to plane is $p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$, giving $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2}$. The centroid of tetrahedron $OABC$ is $G = (\frac{a}{4}, \frac{b}{4}, \frac{c}{4})$. Setting $x = \frac{a}{4}$, $y = \frac{b}{4}$, $z = \frac{c}{4}$ gives $a = 4x$, $b = 4y$, $c = 4z$. Substituting into the distance constraint: $\frac{1}{16x^2} + \frac{1}{16y^2} + \frac{1}{16z^2} = \frac{1}{p^2}$, which simplifies to $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{16}{p^2}$. Comparing with the given locus $x^{-2} + y^{-2} + z^{-2} = 2qp^{-2}$, we get $2q = 16$, so $q = 8$ and $\sqrt{q} = 2\sqrt{2}$.
Correct Answer: 2√2