The answer is no. Here's an example, notice that the varieties are smooth and they intersect pairwise transversally. Consider $X = \mathbb{A}^3$ and set $Z_1 = V(y,z)$, $Z_2 = V(x,z)$, $Z_3 = V(x, z-1)$. Notice that $Z_3$ doesn't intersect any of the other subschemes and so the claim is ok.
Then, $I_1 \cap I_2 = (z, xy)$.
However, $I_1 \cdot I_2 = (xy, yz, xz, z^2)$.
These ideals are not equal clearly. Now, we can immediately see that multiplying/intersecting by $I_3$ won't change the behavior at the origin at all since the ideal doesn't vanish there, so they are not equal. However, just to be sure, I also did the following computation (with Macaulay2):
$$I_1 \cdot I_2 \cdot I_3 = (xz, yz, z^3 - z^2, yz^2 - yz, xz^2 - xz).$$
$$I_1 \cap I_2 \cap I_3 = (z^2 -z, xz, xy).$$
Macaulay2 also confirmed that the ideals were not equal.