Derive the equation of the locus of a point twice as far from (-2,3,4) as from (3,-1,-2)

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{A}\left(-2,3,4\right) \mathrm{and} \mathrm{B}\left(3,-1,-2\right) \\ \mathrm{Let} \mathrm{P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right) \mathrm{be} \mathrm{point} \mathrm{whose} \mathrm{locus} \mathrm{is} \mathrm{to} \mathrm{be} \mathrm{found} \\ \mathrm{We} \mathrm{have} \\ \mathrm{PA}=2\mathrm{PB} \\ \Rightarrow {\mathrm{PA}}^2=4{\mathrm{PB}}^2 \\ \Rightarrow {\left(\mathrm{x}+2\right)}^2+{\left(\mathrm{y}-3\right)}^2+{\left(\mathrm{z}-4\right)}^2=4\left[{\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}+1\right)}^2+{\left(\mathrm{z}+2\right)}^2\right] \\ \Rightarrow {\mathrm{x}}^2+4\mathrm{x}+4+{\mathrm{y}}^2-6\mathrm{y}+9+{\mathrm{z}}^2+16-8\mathrm{z} \\ =4\left[{\mathrm{x}}^2+9-6\mathrm{x}+{\mathrm{y}}^2+2\mathrm{y}+1+{\mathrm{z}}^2+4\mathrm{z}+4\right] \\ \Rightarrow 3{\mathrm{x}}^2+3{\mathrm{y}}^2+3{\mathrm{z}}^2-22\mathrm{x}+14\mathrm{y}+24\mathrm{z}+27=0 \end{gathered}$$

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