Question

$P$ is a variable point which moves such that $3PA=2PB$. If $A=(-2,2,3)$ and $B(13,-3,13)$ prove that $P$ satisfies the equation $x^2+y^2+z^2+28x-12y+10z-247=0$

Answer 1

Given, $A(-2,2,3),B(13,-3,13),P(x,y,z)$

$3PA=2PB$

$\Rightarrow (3PA{)}^2=(2PB{)}^2$

$\Rightarrow 9P{A}^2=4P{B}^2$

$\Rightarrow 9\left[\right(x+2{)}^2+(y-2{)}^2+(z-3{)}^2]=4[(x-13{)}^2+(y+3{)}^2+(z-13{)}^2]$

$\Rightarrow x^2+y^2+z^2+28x-12y+10z-247=0$