Example 2 Given $a b c \neq 0$, and $\frac{a+b}{c}=\frac{b+c}{a}=$ $\frac{c+a}{b}=p$. Then, the line $y=p x+p$ must pass through ( ).
(A) the first and second quadrants
(B) the second and third quadrants
(C) the third and fourth quadrants
(D) the first and fourth quadrants
(1998, National Junior High School Mathematics Party Competition)
Solution: From $\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=p$, we get $a+b=c p, b+c=a p, c+a=b p$.
Adding the three equations, we get
$$
2(a+b+c)=p(a+b+c) .
$$
(1) When $a+b+c=0$, $p=-1, y=-x-1$, then the line passes through the second, third, and fourth quadrants;
(2) When $a+b+c \neq 0$, $p=2, y=2 x+2$, then the line passes through the first, second, and third quadrants.

In summary, the line must pass through the second and third quadrants, so the answer is (B).