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

Let $P=(x,y,z)$

$\text{P}A=\sqrt{(x-(-2){)}^2+(y-2{)}^2+(z-3{)}^2}$

$\text{P}B=\sqrt{(x-13{)}^2+(y-(-3){)}^2+(z-13{)}^2}$

$\text{given that}$ $3PA=2PB$

$3\sqrt{(x+3{)}^2+(y-2{)}^2+(z-3{)}^2}=2\sqrt{(x-13{)}^2+(y+3{)}^2+(z-13{)}^2}$

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

on solving this we get $x^2+y^2+z^2+28x-12y+10z-247=0$