7. A die (a uniform cube with faces numbered $1,2,3,4$, 5,6) is rolled three times. What is the probability that the three numbers on the top faces can form the side lengths of a triangle whose perimeter is divisible by 3? $\qquad$
7. $\frac{11}{72}$.

Let the numbers obtained from rolling the three dice be $a$, $b$, and $c$ ($a \leqslant b \leqslant c$). Then the condition is equivalent to
$$
a+b>c \text {, and } 31(a+b+c) \text {. }
$$

If the triangle formed is equilateral, there are 6 cases.

If the triangle formed is isosceles (but not equilateral), it is easy to prove that $a \equiv b \equiv c(\bmod 3)$. Thus,
$$
(a, b, c)=(1,4,4),(2,5,5),(3,6,6) \text {. }
$$

In this case, there are 9 cases.
If the triangle formed is a scalene triangle (not isosceles), then $ac-b \geqslant 1$. Therefore, $4 \leqslant c \leqslant 6$. Hence,
$$
(a, b, c)=(2,3,4),(3,4,5),(4,5,6) \text {. }
$$

In this case, there are $3 \mathrm{~A}_{3}^{3}=18$ cases.
Therefore, the required probability is $\frac{6+9+18}{6^{3}}=\frac{11}{72}$.