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Q.
Determine the point in $yz$-plane which is equidistant from three points $A(2,0,3)$, $B(0,3,2)$ and $C(0,0,1)$.
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 $PA = PB = PC$.
Now $PA^2 = PB^2$
$\Rightarrow \left(0- 2\right)^{2} + \left( y - 0\right)^{2} + \left(z -3 \right)^{2}$
$= \left( 0 - 0\right)^{2} + \left(y -3 \right)^{2 }+ \left(z -2 \right)^{2}$
i.e., $z - 3y = 0\quad ...\left(1\right)$
and $PB^{2} = PC^{2}$
$\Rightarrow y^{2} + 9 -6y + z^{2} + 4 -4z$
$=y^{2 }+ z^{2}+ 1 - 2z$
i.e., $3y + z = 6 \quad...\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get $y = 1$, $z = 3$
Hence, the coordinates of the point $P$ are $\left(0, 1, 3\right)$.