One. (20 points) Let positive integers $a, b, c (a \geqslant b \geqslant c)$ be the lengths of the sides of a triangle, and satisfy
$$
a^{2}+b^{2}+c^{2}-a b-a c-b c=13 \text {. }
$$

Find the number of triangles that meet the conditions and have a perimeter not exceeding 30.
Given the known equation:
$$
(a-b)^{2}+(b-c)^{2}+(a-c)^{2}=26 \text {. }
$$

Let $a-b=m, b-c=n$.
Then $a-c=m+n(m, n$ are natural numbers $)$.
Thus, equation (1) becomes
$$
m^{2}+n^{2}+m n=13 \text {. }
$$

Therefore, the pairs $(m, n)$ that satisfy equation (2) are:
$$
(m, n)=(3,1),(1,3) \text {. }
$$
(1) When $(m, n)=(3,1)$,
$$
b=c+1, a=b+3=c+4 \text {. }
$$

Since $a, b, c$ are the lengths of the sides of a triangle,
$$
b+c>a \Rightarrow(c+1)+c>c+4 \Rightarrow c>3 \text {. }
$$

Also, the perimeter of the triangle does not exceed 30, i.e.,
$$
\begin{array}{l}
a+b+c=(c+4)+(c+1)+c \leqslant 30 \\
\Rightarrow c \leqslant \frac{25}{3} .
\end{array}
$$

Thus, $3<c \leqslant \frac{25}{3}$.
Therefore, $c$ can take the values $4, 5, 6, 7, 8$, corresponding to 5 triangles that meet the conditions.
(2) When $(m, n)=(1,3)$, similarly,
$1<c \leqslant \frac{23}{3}$.
Thus, $c$ can take the values $2, 3, 4, 5, 6, 7$, corresponding to 6 triangles that meet the conditions.

In summary, the number of triangles that meet the conditions and have a perimeter not exceeding 30 is 11.