Let $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ and $p(x, y)$ be points on $x y$ plane.
Now,
$$
P A^2+P B^2=2 P C^2
$$
$$
\begin{gathered}
\Rightarrow\left(x-x_1\right)^2+\left(y-y_1\right)^2+\left(x-x_2\right)^2+\left(y-y_2\right)^2 \\
=2\left[\left(x-x_3\right)^2+\left(y-y_3\right)^2\right] \\
\Rightarrow x^2+x_1^2-2 x x_1+y^2+y_1^2-2 y y_1 \\
\quad+x^2+x_2^2-2 x x_2+y^2+y_2^2-2 y y_2 \\
=2\left[x^2+x_3^2-2 x x_3+y^2+y_3^2-2 y y_3\right] \\
\Rightarrow x\left(-2 x_1-2 x_2+4 x_3\right)+y\left(-2 y_1-2 y_2+4 y_3\right) \\
\quad+x_1^2+x_2^2+y_1^2+y^2-2 x_3^2-2 y_3^2=0
\end{gathered}
$$
this equation is in the form of
$$
a x+b y+c=0
$$
Hence, locus of $p$ is a straight line.