7. $\frac{11}{72}$.

Let the numbers obtained from rolling the three dice be $a, b, c$, then $a, b, c \in\{1,2,3,4,5,6\}$. Without loss of generality, assume $a \leqslant b \leqslant c$, then the condition is equivalent to $a+b>c$ and $3 \mid(a+b+c)$.
If the triangle formed is equilateral, there are 6 cases.
If the triangle formed is isosceles (but not equilateral), it is not difficult to prove that $a \equiv b \equiv c(\bmod 3)$. If $a=b+3$, this requires $a>3$, thus $c>6$, which is a contradiction. If $a<b=c$, then $b=c=a+3$, thus $b=c \geqslant 4$, and $(a, b, c)$ can be $(1,4,4),(2,5,5),(3,6,6)$, in which case there are $3 \times \frac{\mathrm{A}_{3}^{3}}{\mathrm{~A}_{2}^{2}}=9$ cases.

If the triangle formed is a scalene triangle (not isosceles), then $a<b<c$, thus $a \geqslant 2$ (otherwise $a+b=b+1 \leqslant c$, which is a contradiction), and $4 \leqslant c \leqslant 6$.
When $c=4$, it is found that only $(a, b)=(2,3)$ satisfies the condition;
When $c=5$, it is found that only $(a, b)=(3,4)$ satisfies the condition;
When $c=6$, it is found that only $(a, b)=(4,5)$ satisfies the condition.
In this case, there are $3 \times \mathrm{A}_{3}^{3}=18$ cases.
Therefore, the required probability is $\frac{6+9+18}{6^{3}}=\frac{11}{72}$.
7. If a die is thrown three times in succession, the probability that the three numbers that appear can form the side lengths of a triangle whose perimeter is divisible by 3 is $\qquad$ . .