Let's find a triangle where the lengths of the sides are integers and their product is equal to 600, and the perimeter is as small as possible. Solve the same problem if the product is 144.
I. If a triangle has side lengths that are integers, and their product is 600, then none of the sides can be 1 unit, because then the other two sides would have to be equal to make their difference less than 1, but 600 cannot be factored into the product of two equal integers, as it is not a perfect square.

However, if each side is at least 2, then the product of any two sides is not less than twice the larger side, and thus not less than the sum of the two sides, which is greater than the third side; in this case, the product of the three sides, which is 600, is greater than the square of the third side. Since we could start from any two sides, the square of each side is less than 600, so it is not greater than 24.

Factoring 600 into prime factors, we get $2^{3} \cdot 3 \cdot 5^{2}$. There must be two sides whose lengths are divisible by 5, because no side length can be divisible by $5^{2}$, which is greater than 24. These two sides are either equal or differ by at least 5, so the length of the third side is at least 6.

The 600 has factorizations into three factors where two equal factors are divisible by 5: $5 \cdot 5 \cdot 24$ and $10 \cdot 10 \cdot 6$. The factors of the first cannot be the side lengths of a triangle $(5+5<24)$, but the factors of the second can. Among the factorizations where two different factors are divisible by 5 and the third factor is greater than their difference, only $5 \cdot 10 \cdot 12$ exists, and its factors can also be the side lengths of a triangle.

Among the triangles formed by the side lengths 10, 10, 6 and 5, 10, 12, the perimeter of the former is smaller, so the sides of the sought triangle are 10, 10, and 6 units.

II. In the second case, $a b c=144$. We can reason similarly to the above. 144 is a perfect square, and a triangle can be constructed from the side lengths $1,12,12$.

If the smallest side is 2, the product of the other two is 72, which is not a perfect square, so their difference is 1, and the factors are 8 and 9.

If the smallest side is 3, the product of the other two is 48, which is again not a perfect square, so their difference is 1 or 2, and only the $6 \cdot 8$ factorization fits.

Finally, if the smallest side is 4, the product of the other two sides is 36, and their difference is 0, 1, 2, or 3, here only the $6 \cdot 6$ factorization fits. - The found

$$
1,12,12 ; \quad 2,8,9 ; \quad 3,6,8 \quad \text { and } 4,6,6
$$

Among these triangles, the last one has the smallest perimeter compared to the other three, so the sides are 4, 6, and 6 units.