Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Given $\frac{abc}{6}=32$ where $A(a,0,0),B(0,b,0),C(0,0,c)$. (A) Centroid $(\alpha ,\beta ,\gamma )=\left(\frac{a}{4},\frac{b}{4},\frac{c}{4}\right)\Rightarrow a=4\alpha$ etc.; $64\alpha \beta \gamma =192\Rightarrow \alpha \beta \gamma =3:xyz=3\to \mathrm{R}$. (B) Equidistant point $(\alpha ,\beta ,\gamma )=\left(\frac{a}{2},\frac{b}{2},\frac{c}{2}\right):8\alpha \beta \gamma =192\Rightarrow \alpha \beta \gamma =24:xyz=24$ → P. (C) Foot of perpendicular from origin to the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ is $(\frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{\alpha }$-form $):a=\frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{\alpha }$ etc.; $abc=192\Rightarrow {\left({\alpha }^2+{\beta }^2+{\gamma }^2\right)}^3=192\alpha \beta \gamma \to \mathrm{Q}$. (D) $PA\perp PB\Rightarrow a\alpha +b\beta ={\alpha }^2+{\beta }^2+{\gamma }^2$ and $PB\perp PC\Rightarrow b\beta +c\gamma ={\alpha }^2+{\beta }^2+{\gamma }^2$, giving $a=\frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{2\alpha }$ etc.; $abc=192\Rightarrow {\left({\alpha }^2+{\beta }^2+{\gamma }^2\right)}^3=192\times 8\alpha \beta \gamma =1536\alpha \beta \gamma \to \mathrm{S}$.