Let $(a, b, c, d)$ be a set of positive integers that satisfy the conditions. Then, from (3), we have $b > a$, $d > \frac{bc}{b-a}$.

Consider the following quadratic function:
$$\begin{aligned} f(x) & = (9ac + bx)(ax + bc) - a^2x^2 - 10abcx - b^2c^2 \\ & = a(b-a)x^2 + c(9a-b)(a-b)x + bc^2(9a-b) \end{aligned}$$

The coefficient of the quadratic term is $a(b-a) > 0$, and $f(d) = 0$.
Notice that the discriminant of the function $f(x)$ is
$$\Delta = -(9a-b)(b-3a)^2c^2$$

Therefore, if $b < 3a$, then $\Delta > 0$, which contradicts $f(d) = 0$.

If $b \geqslant 9a$, then the axis of symmetry of the quadratic function $f(x)$ is $x = \frac{c(9a-b)}{2a} < \frac{bc}{b-a} > 0$, so
$$f(d) > f\left(\frac{bc}{b-a}\right) = \frac{ab^2c^2}{b-a} > 0$$

which also contradicts $f(d) = 0$.
Thus, it can only be that $b = 3a$. In this case, $\Delta = 0$, so $f(d) = 0$ is the minimum value of the function $f(x)$, hence $d = \frac{c(9a-b)}{2a}$, i.e., $d = 3c$.

Direct verification shows that when $a, c \in \mathbf{N}^{*}$, the array $(a, 3a, c, 3c)$ is a set of positive integers that meet the requirements, and these are the answers we seek.