Exactly two of the six numbers are multiples of $3$ and these two need to be multiplied together, otherwise two of the three terms of the equality are multiples of $3$ but the third one is not.
Let $n$ and $n+3$ denote these multiples of $3$. Two of the four remaining numbers give remainder $1$ when divided by $3$, while the other two give remainder $2$, so the two other products are either $\equiv 1 \pmod{3}$ and $\equiv 2 \pmod{3}$, or they are both $\equiv 1 \pmod{2}$ and $\equiv 2 \pmod{3}$. In conclusion, the term $n(n+3)$ needs to be on the right hand side of the equality.
Looking at parity, three of the numbers are odd, and three are even. One of $n$ and $n+3$ is odd, the other even, so exactly two of the other numbers are odd. As $n(n+3)$ is even, the two remaining odd numbers need to appear in different terms.
We distinguish the following cases:
I. The numbers are $n-2, n-1, n, n+1, n+2, n+3$.

The product of the two numbers on the RHS needs to be larger than $n(n+3)$. The only possibility is $(n-2)(n-1)+n(n+3)=(n+1)(n+2)$ which leads to $n=3$. Indeed,
$1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5.$

II. The numbers are $n-1, n, n+1, n+2, n+3, n+4$.
As $(n+4)(n-1)+n(n+3)=(n+1)(n+2)$ has no solutions, $n+4$ needs to be on the RHS, multiplied with a number having a different parity, so $n-1$ or $n+1$. $(n+2)(n-1)+n(n+3)=(n+1)(n+4)$ leads to $n=3$. Indeed, $2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7$.
$$(n+2)(n+1)+n(n+3)=(n-1)(n+4) \text{ has no solution.}$$
III. The numbers are $n, n+1, n+2, n+3, n+4, n+5$.
We need to consider the following situations: $(n+1)(n+2)+n(n+3)=(n+4)(n+5)$ which leads to $n=6$; indeed $7 \cdot 8 + 6 \cdot 9 = 10 \cdot 11$; $(n+2)(n+5)+n(n+3)=(n+1)(n+4)$ obviously without solutions, and $(n+1)(n+4)+n(n+3)=(n+2)(n+5)$ which leads to $n=2$ (not a multiple of $3$).

In conclusion, the problem has three solutions:
$$1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5, \quad 2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7 \quad \text{and} \quad 7 \cdot 8 + 6 \cdot 9 = 10 \cdot 11.$$