Let's determine the positive integers $a, b$ for which

$$
(\sqrt{30}-\sqrt{18})(3 \sqrt{a}+\sqrt{b})=12
$$
Divide both sides of the equality by $(\sqrt{30}-\sqrt{18})$, then rationalize the denominator of the resulting fraction:

$$
3 \sqrt{a}+\sqrt{b}=\frac{12}{\sqrt{30}-\sqrt{18}}=\frac{12(\sqrt{30}+\sqrt{18})}{30-18}
$$

from which

$$
3 \sqrt{a}+\sqrt{b}=\sqrt{30}+\sqrt{18}=3 \sqrt{2}+\sqrt{30}
$$

follows.

We will show that if $a$ and $b$ are positive integers, then (2) can only hold if the terms on both sides of the equation are equal, i.e., $a=2$ and $b=30$. Square both sides of (2). Since both sides of (2) are positive, the resulting equation is equivalent to the original.

$$
9 a+b+6 \sqrt{a b}=48+6 \sqrt{60}
$$

from which we get

$$
\frac{48-(9 a+b)}{6}=\sqrt{a b}-\sqrt{60}
$$

According to the condition, the left side is a ratio of integers, so $\sqrt{a b}-\sqrt{60}$ is a rational number. If it is not zero, then

$$
\frac{a b-60}{\sqrt{a b}-\sqrt{60}}=\sqrt{a b}+\sqrt{60}
$$

is also rational, and thus $2 \sqrt{60}=(\sqrt{a b}+\sqrt{60})-(\sqrt{a b}-\sqrt{60})$ is also rational. However, this is not true, since it is known that among positive integers, only the square roots of perfect squares are rational, and 60 is not a perfect square.

Thus, we have that $\sqrt{a b}-\sqrt{60}=0$, from which we get, on the one hand,

$$
a b=60, \text { and on the other hand, according to (3), }
$$

$$
9 a+b-48=0
$$

The solutions of the system of equations are $\left(\frac{10}{3}, 18\right)$ and $(2,30)$. Since only the latter solution consists of integers, indeed $a=2$ and $b=30$, as we claimed.

Remark. In the solution, we referred to the fact that if a positive integer $m$ is not a perfect square, then $\sqrt{m}$ is irrational. This can be shown by a similar line of reasoning to the well-known proof of the irrationality of $\sqrt{2}$. 

If $\sqrt{m}$ is rational, then it can be written as the ratio of relatively prime integers $p / q$. Squaring and multiplying by $q^{2}$, we get $p^{2}=m \cdot q^{2}$. From this, on the one hand, $m \mid p^{2}$, and on the other hand, since by the fundamental theorem of arithmetic $p^{2}$ and $q^{2}$ have no common divisor greater than 1, all prime factors of $p^{2}$ are in the factorization of $m$, so $p^{2} \mid m$. Combining the two divisibilities, we indeed get $m=p^{2}$, i.e., $m$ is a perfect square.

The above proof is not limited to square roots; it is generally true that the $n$-th root of a natural number is either an integer (and thus the number is the $n$-th power of that integer) or irrational.