Solution:
Using the standard notation for the elements of $\triangle ABC$ we have $b^{2} = l_{a}^{2} = bc - \frac{a^{2}bc}{(b+c)^{2}}$. Hence
$$a^{2}c = (c-b)(c+b)^{2}$$
Let $\frac{a}{c} = \frac{m}{n},\ (m, n) = 1$, and $\frac{b}{c} = \frac{r}{s},\ (r, s) = 1$. Then (1) implies that
$$\frac{m^{2}}{n^{2}} = \frac{(s-r)(s+r)^{2}}{s^{3}}$$
Since both sides are irreducible fractions, we obtain $m^{2} = (s-r)(s+r)^{2}$ and $n^{2} = s^{3}$. The first equality shows that $s-r$ is a perfect square and the second one implies that $s$ is a perfect square. Set $s = t^{2}$ and $s-r = k^{2}$. Then $r = t^{2} - k^{2}$, $m = k\left(2t^{2} - k^{2}\right)$ and $n = t^{3}$.

We now set $a = mx$, $c = nx$, $b = ry$ and $c = sy$. Then $nx = sy$, i.e. $y = tx$. Therefore $a = xk\left(2t^{2} - k^{2}\right)$, $b = x t\left(t^{2} - k^{2}\right)$ and $c = x t^{3}$, where $t > k$ and $(t, k) = 1$. Moreover, one can easily check that these $a, b$ and $c$ satisfy the triangle inequality.

Now the condition $a + b + c = 10p$ becomes $x(k + t)\left(2t^{2} - k^{2}\right) = 10p$. Note that $\left(k + t, 2t^{2} - k^{2}\right) = 1$. Then it is easy to see that we have only the following possibilities:
$$\begin{array}{l|l|l} x = 1 \\ k + t = 5 \\ 2t^{2} - k^{2} = 2p \end{array}, \quad \begin{aligned} & x = 1 \\ & k + t = 10 \\ & 2t^{2} - k^{2} = p \end{aligned}, \quad \begin{aligned} & x = 2 \\ & k + t = 5 \\ & 2t^{2} - k^{2} = p \end{aligned}$$
A direct verification shows that $(x, k, t) = (1, 2, 3)$, $(x, k, t) = (1, 3, 7)$ and $(x, k, t) = (2, 1, 4)$. Therefore $(a, b, c) = (28, 15, 27)$, $(a, b, c) = (267, 280, 343)$ and $(a, b, c) = (62, 120, 128)$.