Answer:
Y plane at y=3/2 contains all the points which will be equidistant from both points P and Q.
Step-by-step explanation:
As the complete question is not readable, the question is found online and is attached herewith.
By the distance formula, as the point R with coordinates x,y,z is equidistant from P and Q thus
[tex]|PR|=|QR|\\\sqrt{(x-0)^2+(y-0)^2+(z-0)^2}=\sqrt{(x-0)^2+(y-3)^2+(z-0)^2}\\(x-0)^2+(y-0)^2+(z-0)^2=(x-0)^2+(y-3)^2+(z-0)^2\\x^2+y^2+z^2=x^2+(y-3)^2+z^2\\y^2=(y-3)^2\\y^2=y^2+9-6y\\-6y=-9\\y=3/2[/tex]
So the y plane at y=3/2 contains all the points which will be equidistant from both points P and Q.
