CGPET 2015 Solved Paper

Let the plane meets the axes at A(a, 0, 0), B(0, b, 0) and C(0, 0, c). Then, equation of plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ It is given that plane is at a constant distance p from the origin (0, 0, 0). ∴$p = \left| \frac{0+0-1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} \right|$ ⇒$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$ ..(i) Let (α, β, γ)be the coordinates of centroid of the formed tetrahedron. Then, $\alpha = \frac{a+0+0+0}{4} \Rightarrow a = 4\alpha$ $\beta = \frac{0+b+0+0}{4} \Rightarrow b = 4\beta$ $\gamma = \frac{0+0+c+0}{4} \Rightarrow c = 4\gamma$ On putting values of a, b, c in Eq. (i), we get $\frac{1}{p^2} = \frac{1}{(4\alpha)^2} + \frac{1}{(4\beta)^2} + \frac{1}{(4\gamma)^2}$ ⇒ $\frac{16}{p^2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2}$ ∴ Locus of centroid of tetrahedron is $\frac{16}{p^2} = \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}$