Example 6 Let $a, b, c$ be three distinct real numbers, and $p(x)$ be a polynomial with real coefficients. It is known that: (1) $p(x)$ divided by $x-a$ yields a remainder of $a$; (2) $P(x)$ divided by $x-b$ yields a remainder of $b$; (3) $p(x)$ divided by $x-c$ yields a remainder of $c$. Find the remainder when the polynomial $p(x)$ is divided by $(x-a)(x-b)(x-c)$.
(1990 Italian Competition Problem)
Given that the remainder of $p(x)$ divided by $x-a$ is $a$, we can set
$$
p(x)=(x-a) \cdot Q_{1}(x)+a \text {. }
$$

Thus, $p(b)=(b-a) \cdot Q_{1}(b)+a=b$. Noting that $a \neq b$,
we have $Q_{1}(b)=1$, which means $Q_{1}(x)=(x-b) \cdot Q_{2}(x)+1$.
Substituting the above into (1), we get $p(x)=(x-a)(x-b) \cdot Q_{2}(x)+x$.
From condition (3), we have $p(c)=(c-a)(c-b) \cdot Q_{2}(c)+c=c$,
which gives $Q_{2}(c)=0$, so $Q_{2}(x)=(x-c) \cdot Q_{3}(x)$.
Substituting the above into (2), we get $p(x)=(x-a)(x-b)(x-c) \cdot Q_{3}(x)+x$.
Therefore, the remainder of $p(x)$ divided by $(x-a)(x-b)(x-c)$ is $x$.