Problem 798

Solution 1: Clearly, $a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\left(14-a_{3}\right)=14$, where $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3, 14-a_{3} \geqslant 0$. Transform the equation to:
$$\left(a_{1}-1\right)+\left(a_{2}-a_{1}-3\right)+\left(a_{3}-a_{2}-3\right)+\left(14-a_{3}\right)=7.$$

At this point, $a_{1}-1, a_{2}-a_{1}-3, a_{3}-a_{2}-3, 14-a_{3} \geqslant 0$. The above indeterminate equation has $C_{10}^{7}=C_{10}^{3}$ different non-negative integer solutions. Therefore, the number of different ways to meet the requirements is $C_{10}^{3}=120$.

Solution 2: Let $S=\{1,2, \cdots, 14\}, S^{\prime}=\{1,2, \cdots, 10\}, \left\{a_{1}, a_{2}, a_{3}\right\}$ be a three-element subset of $S$, and $\left\{a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}\right\}$ be a three-element subset of $S^{\prime}$, satisfying $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-2, a_{3}^{\prime}=a_{3}-4$, i.e., establish the following correspondence:
$$\left(a_{1}, a_{2}, a_{3}\right) \rightarrow\left(a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}\right)=\left(a_{1}, a_{2}-2, a_{3}-4\right).$$

Clearly, this is a one-to-one correspondence from $S$ to $S^{\prime}$. Thus, the number of ways to choose is equal to the number of ways to choose any three different numbers from $S^{\prime}$, which is $C_{10}^{3}=120$.

Solution 3: Construct the following model:
Take 10 identical white balls and arrange them in a row. Take 5 different black balls and divide them into 3 groups in the order of $2,2,1$, then insert them into the 10 gaps between the white balls, from left to right, excluding the left end but including the right end. The number of insertion methods is $C_{10}^{3}$. Each insertion method corresponds to an ordered array, as shown in the figure:
Thus, the number of different methods is $C_{10}^{3}=120$.