(3) Given that $P$ is any point inside triangle $A B C$ (excluding the boundary), and satisfies $(\overrightarrow{P B}-$ $\overrightarrow{P A}) \cdot(\overrightarrow{P B}+\overrightarrow{P A}-2 \overrightarrow{P C})=0$, then $\triangle A B C$ must be ( ).
(A) Right triangle
(B) Equilateral triangle
(C) Isosceles triangle
(D) Isosceles right triangle
(3) Since
$$
\begin{array}{c}
\overrightarrow{P B}-\overrightarrow{P A}=\overrightarrow{A B}, \\
\overrightarrow{P B}+\overrightarrow{P A}-2 \overrightarrow{P C}=\overrightarrow{C B}+\overrightarrow{C A},
\end{array}
$$

the given conditions can be rewritten as
$$
\overrightarrow{A B} \cdot(\overrightarrow{C B}+\overrightarrow{C A})=0 \text {. }
$$

It is easy to see that this triangle is an isosceles triangle. Therefore, the answer is C.