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Q.
The point in YZ-plane which is equidistant from three points $A (2,0,3), B (0,3,2)$ and $C (0,0,1)$ is
Introduction to Three Dimensional Geometry
Solution:
Since x-coordinate of every point in YZ-plane is zero.
Let $P (0, y , z )$ be a point on the YZ-plane such that
$PA = PB = PC . \text { Now, } PA = PB$
$\Rightarrow (0-2)^{2}+( y -0)^{2}+( z -3)^{2}$
$=(0-0)^{2}+( y -3)^{2}+( z -2)^{2}$
i.e., $z -3 y =0 \text { and } PB = PC$
$\Rightarrow y ^{2}+9-6 y + z ^{2}+4-4 z = y ^{2}+ z ^{2}+1-2 z,$
i.e., $3 y + z =6$
On simplifying the two equations, we get $y =1$ and $z =3$. Here, the coordinate of the point $P$ are $(0,1,3)$.