3. If three numbers are taken simultaneously from the 14 integers $1,2, \cdots, 14$, such that the absolute difference between any two numbers is not less than 3, then the number of different ways to choose is
3. 120.

Let the three integers taken out be $x, y, z (x<y<z)$.
Let $a=x, b=y-x-2$,
$$
c=z-y-2, d=15-z \text {. }
$$

Thus, $a, b, c, d \geqslant 1$.
If $a, b, c, d$ are determined, then $x, y, z$ are uniquely determined.
Since $a+b+c+d=11$, it is equivalent to dividing 11 identical balls into four piles.
Using the stars and bars method, we get $\mathrm{C}_{10}^{3}=120$.