Question
Find the locus of $P$ if $P A^2+P B^2=2 k^2$, where $A$ and $B$ are the points $(3,4,5)$ and $(-1,3,-7)$.

Answer

Let $\mathrm{P}(\mathrm{x} 1, \mathrm{y} 1, \mathrm{z})$,
Here, $\mathrm{A}(3,4,5), \mathrm{B}(-1,3,-7) \mathrm{PA}^2+\mathrm{PB}^2=2 \mathrm{k}^2$
$\Rightarrow(\mathrm{x}-3)^2+(\mathrm{y}-4)^2+(\mathrm{z}-5)^2+(\mathrm{x}+1)^2+(\mathrm{y}-3)^2+(2+7)^2$
$=2 k^2$
$\Rightarrow x^2+9-6 x+y^2+16-8 y+z^2+25-10 z+x^2+1+2 x+y^2+9-6 y+z^2+49+14 z=2 k^2$
$\Rightarrow 2 x^2+2 y^2+2 z^2-4 x-14 y+4 z+109=2 k^2$
$\Rightarrow 2\left(x^2+y^2+z^2\right)-4 x-14 y+4 z+109-2 k^2=0$

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