Answer
$y=\frac{3}{2}$
Work Step by Step
Use the distance formula $(x-0)^2+(y-0)^2+(z-0)^2=d^2=(x-0)^2+(y-3)^2+(z-0)^2$ or
$x^2+y^2+z^2=x^2+y^2-6y+9+z^2$ so we have $y=\frac{3}{2}$ which represents a plane parallel to the xz-plane.
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 9 - Section 9.3 - Three-Dimensional Coordinate Geometry - 9.3 Exercises - Page 653: 23
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