5. If $k=\frac{2 a+b}{c}=\frac{2 c+a}{b}=\frac{2 b+c}{a}$, then the value of $k$ is ( ).
(A) 1
(B) 2
(C) 3
(D) None of the above
5. (C).

Since $k=\frac{2 a+b}{c}=\frac{2 c+a}{b}=\frac{2 b+c}{a}$, we have
$$
\begin{array}{l}
2 a b+b^{2}=2 c^{2}+a c, \\
2 a c+a^{2}=2 b^{2}+b c, \\
2 b c+c^{2}=2 a^{2}+a b .
\end{array}
$$

Adding the three equations, we get
$$
a b+b c+a c=a^{2}+b^{2}+c^{2},
$$

Multiplying both sides by 2, rearranging, and completing the square, we get
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0 .
$$

Therefore, $a-b=b-c=c-a=0$, so $a=b=c$.
Thus, $k=\frac{2 a+b}{c}=\frac{3 c}{c}=3$.