214. One of the angles of a triangle is $120^{\circ}$, and the lengths of the sides form an arithmetic progression. Find the ratio of the lengths of the sides of the triangle.
$\triangle$ Let the lengths of the sides of the triangle in ascending order be $a$, $a+d$, and $a+2d$. Then the largest side is opposite the largest angle of the triangle, which is $120^{\circ}$.

By the Law of Cosines, we have:

$$
\begin{aligned}
& (a+2 d)^{2}=a^{2}+(a+d)^{2}-2 a(a+d) \cos 120^{\circ} \\
& a^{2}+4 a d+4 d^{2}=a^{2}+a^{2}+2 a d+d^{2}+a^{2}+a d \\
& 2 a^{2}-a d-3 d^{2}=0, \quad 2\left(\frac{a}{d}\right)^{2}-\frac{a}{d}-3=0
\end{aligned}
$$

From this,

$$
\frac{a}{d}=\frac{3}{2}, \quad a=\frac{3}{2} d, \quad a+d=\frac{5}{2} d, \quad a+2 d=\frac{7}{2} d
$$

60

Therefore, the ratio of the lengths of the sides of the triangle is

$$
a:(a+d):(a+2 d)=\frac{3}{2} d: \frac{5}{2} d: \frac{7}{2} d=3: 5: 7
$$

Answer: $3: 5: 7$.