A variable plane makes with the coordinate planes, a tetrahedron of constant volume 64k^​3. Then find the locus of the centroid of the tetrahedron. Please answer this as quick as possible.

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$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{variable} \mathrm{plane} \mathrm{be} \mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0 \\ \mathrm{Now} \\ \mathrm{Let} \mathrm{plane} \mathrm{meets} \mathrm{X},\mathrm{Y},\mathrm{Z} \mathrm{axes} \mathrm{at} \mathrm{A},\mathrm{B},\mathrm{C} \mathrm{respectively} \mathrm{and} \mathrm{O} \mathrm{be} \mathrm{the} \mathrm{origin}. \\ \mathrm{Hence} \mathrm{vertices} \mathrm{are} \mathrm{tertahedron} \mathrm{are} \mathrm{O},\mathrm{A},\mathrm{B},\mathrm{C}. \\ \mathrm{We} \mathrm{have} \mathrm{O}\left(0,0\right) \\ \mathrm{At} \mathrm{X} \mathrm{axis}, \mathrm{y}=0, \mathrm{z}=0 \\ \Rightarrow \mathrm{ax}+0+0+\mathrm{d}=0 \\ \Rightarrow \mathrm{x}=-\frac{\mathrm{d}}{\mathrm{a}} \\ \mathrm{Hence} \\ \mathrm{Co}-\mathrm{ordinates} \mathrm{of} \mathrm{A} \mathrm{are} \mathrm{A}\left(-\frac{\mathrm{d}}{\mathrm{a}},0,0\right) \\ \mathrm{Similarly} \\ \mathrm{Co}-\mathrm{ordinates} \mathrm{of} \mathrm{B} \mathrm{are} \mathrm{B}\left(0,-\frac{\mathrm{d}}{\mathrm{b}},0\right) \\ \mathrm{Co}-\mathrm{ordinates} \mathrm{of} \mathrm{C} \mathrm{are} \mathrm{C}\left(0,0,-\frac{\mathrm{d}}{\mathrm{c}}\right) \\ \mathrm{Hence} \\ \overrightarrow{\mathrm{OA}}=-\frac{\mathrm{d}}{\mathrm{a}}\hat{\mathrm{i}} \\ \mathrm{Similarly} \\ \overrightarrow{\mathrm{OB}}=-\frac{\mathrm{d}}{\mathrm{b}}\hat{\mathrm{j}} \\ \overrightarrow{\mathrm{OB}}=-\frac{\mathrm{d}}{\mathrm{c}}\hat{\mathrm{k}} \\ \mathrm{Let} \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{centroid} \mathrm{be} \left(\mathrm{p},\mathrm{q},\mathrm{r}\right) \\ \Rightarrow \mathrm{p}=\frac{-\frac{\mathrm{d}}{\mathrm{a}}+0+0+0}{4} \\ \Rightarrow -\frac{\mathrm{d}}{\mathrm{a}}=4\mathrm{p} \\ \mathrm{Similarly} \\ -\frac{\mathrm{d}}{\mathrm{b}}=4\mathrm{q} \\ -\frac{\mathrm{d}}{\mathrm{c}}=4\mathrm{r} \\ \mathrm{Hence} \\ \overrightarrow{\mathrm{OA}}=4\mathrm{p}\hat{\mathrm{i}} \\ \overrightarrow{\mathrm{OB}}=4\mathrm{q}\hat{\mathrm{j}} \\ \overrightarrow{\mathrm{OC}}=4\mathrm{r}\hat{\mathrm{k}} \\ \mathrm{Volume} \mathrm{of} \mathrm{tetrahedron}=64{\mathrm{k}}^3 \\ \Rightarrow \left|\frac{1}{6}\overrightarrow{OA}.\left(\overrightarrow{OB}\times \overrightarrow{OC}\right)\right|=64{\mathrm{k}}^3 \\ \Rightarrow \frac{1}{6}\left(4\mathrm{p}\hat{\mathrm{i}}\right)\left(4\mathrm{q}\hat{\mathrm{j}}\times 4\mathrm{r}\hat{\mathrm{k}}\right)=\pm 64{\mathrm{k}}^3 \\ \Rightarrow \frac{4\times 4\times 4}{6}\left(\mathrm{p}\hat{\mathrm{i}}\right)\left(\mathrm{q}\hat{\mathrm{j}}\times \mathrm{r}\hat{\mathrm{k}}\right)=\pm 64{\mathrm{k}}^3 \\ \Rightarrow \frac{64}{6}\left(\mathrm{p}\hat{\mathrm{i}}\right)\left\{\left(\mathrm{qr}\right)\left(\hat{\mathrm{j}}\times \hat{\mathrm{k}}\right)\right\}=\pm 64{\mathrm{k}}^3 \\ \Rightarrow \left(\mathrm{p}\hat{\mathrm{i}}\right)\left\{\left(\mathrm{qr}\hat{\mathrm{i}}\right)\right\}=\pm 6{\mathrm{k}}^3 \left[\left(\hat{\mathrm{j}}\times \hat{\mathrm{k}}\right)=\hat{\mathrm{i}}\right] \\ \Rightarrow \mathrm{pqr}=\pm 6{\mathrm{k}}^3 \\ \mathrm{To} \mathrm{get} \mathrm{equation} \mathrm{of} \mathrm{locus}, \\ \mathrm{Replace} \mathrm{p}\to \mathrm{x}, \mathrm{q}\to \mathrm{y}, \mathrm{r}\to \mathrm{z} \\ \Rightarrow \boxed{\mathbf{xyz}\mathbf{=}\mathbf{\pm }\mathbf{6}{\mathbf{k}}^{\mathbf{3}}} \end{gathered}$$

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