Given vertices of the triangle $\mathrm{ABC} A(-1,3), \mathrm{B}(3,5)$ and $\mathrm{C}(5,7)$
According to the question, $\mathrm{PA}=\mathrm{PB}=\mathrm{PC}$


$$
\begin{aligned}
& \sqrt{(x+1)^2+(y-3)^2}=\sqrt{(x-3)^2+(y-5)^2}= \\
& \sqrt{(x-5)^2+(y-7)^2}
\end{aligned}
$$
Take square both sides,
$$
\begin{aligned}
& \mathrm{x}^2+1+2 \mathrm{x}+\mathrm{y}^2+9-6 \mathrm{y}=\mathrm{x}^2+9-6 \mathrm{x}+\mathrm{y}^2+25-10 \mathrm{y}=\mathrm{x}^2+10 \mathrm{x}- \\
& +25+\mathrm{y}^2+49-14 \mathrm{y} \\
& \text { compare each terms, } \\
& \mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}-6 \mathrm{y}+10=\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}-10 \mathrm{y}+34=\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}- \\
& 14 \mathrm{y}+74 \\
& \mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}-6 \mathrm{y}+10=\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}-10 \mathrm{y}+34 \\
& 8 \mathrm{x}+4 \mathrm{y}=24
\end{aligned}
$$

$\begin{aligned} & x^2+y^2-6 x-10 y+34=x^2+y^2-10 x-14 y+74 \\ & 4 x+4 y=40\end{aligned}$

subtract (ii) from (i)
$$
\begin{aligned}
& 2 x+y=6 \\
& x+y=10 \\
& x=-4
\end{aligned}
$$
from(i)
$$
\begin{aligned}
& 2 x+y=6 \\
& 2 x-4+y=6 \\
& -8+y=6 \\
& y=14 \\
& \text { so, } p(x, y) \rightarrow(-4,14)
\end{aligned}
$$
Now, $P A=\sqrt{(-4+1)^2+(14-3)^2}=\sqrt{9+121}=\sqrt{130}$
So, option (d) is correct.