Answer
$4x-3y-z=-10$
Work Step by Step
1. Assume point $R(x,y,z)$ is on the plane satisfying the condition given in the problem.
2. Use the distance formula, we have $d^2_{PR}=(x+3)^2+(y-2)^2+(z-5)^2$ and $d^2_{QR}=(x-1)^2+(y+1)^2+(z-4)^2$.
3. Since $d_{PR}=d_{QR}$, we have $(x+3)^2+(y-2)^2+(z-5)^2=(x-1)^2+(y+1)^2+(z-4)^2$ which gives $6x+9-4y+4-10z+25=-2x+1+2y+1-8z+16$ or $8x-6y-2z+20=0$.
4. Thus we obtain the plane equation as $4x-3y-z=-10$
