Thank you for reporting, we will resolve it shortly
Q.
If the point $A (3,2,2)$ and $B (5,5,4)$ are equidistant from $P$, which is on $x$ -axis, then the coordinates of $P$ are
Introduction to Three Dimensional Geometry
Solution:
The point on the $x$ -axis is of the form $P ( x , 0,0)$.
Since, the points $A$ and $B$ are equidistant from $P$.
Therefore, $PA ^{2}= PB ^{2}$
i.e.,$(x-3)^{2}+(0-2)^{2}+(0-2)^{2}$
$=(x-5)^{2}+(0-5)^{2}+(0-4)^{2}$
$\Rightarrow 4 x=25+25+16-17$ i.e., $x=\frac{49}{4}$
Thus, the point $P$ on the $x$ -axis is $\left(\frac{49}{4}, 0,0\right)$ which is equidistant from $A$ and $B$.