We represent the given information by a graph the vertices of which correspond to the $117$ spies. Two vertices are connected iff the corresponding spies share a password which allows them to communicate directly. We need to ensure that at least one of the sub-graphs formed by spies sharing the same mission is connected.

To find the number of edges needed to ensure that a graph with $a \ge 2$ vertices is connected, no matter how the edges are placed, we first observe that the complete graph with $a-1$ vertices has $\frac{(a-1)(a-2)}{2}$ edges. We next prove that it is sufficient to have
$$\frac{(a-1)(a-2)}{2} + 1 = \frac{a^2 - 3a + 4}{2}$$
edges. To see this, assume that the graph is not connected. Then there exist two sub-graphs, not connected to each other, one with $x \ge 1$ and the other with $a-x \ge 1$ vertices. The number of edges in such a graph is at most
$$\begin{aligned} \frac{(a-x)(a-x-1)}{2} + \frac{x(x-1)}{2} &= \frac{2x^2 - 2ax + a^2 - a}{2} = \\ &= \frac{a^2 - 3a + 2}{2} - (a-x-1)(x-1) \le \frac{a^2 - 3a + 2}{2} = \frac{(a-1)(a-2)}{2} \end{aligned}$$

If the sizes of the three missions are $a$, $b$ and $c$, the above shows that issuing
$$n = \frac{a^2 - 3a + 2}{2} + \frac{b^2 - 3b + 2}{2} + \frac{c^2 - 3c + 2}{2} + 1$$
passwords, ensures that at least one mission is networked. Finally, we need to minimise $2n = a^2 + b^2 + c^2 - 3(a+b+c) + 8$ subject to the condition $a+b+c = 117$. Because
$$\begin{aligned} 3(a^2 + b^2 + c^2) - (a+b+c)^2 &= 2(a^2 + b^2 + c^2 - ab - bc - ca) = \\ &= (a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0, \end{aligned}$$
we obtain
$$2n \ge \frac{(a+b+c)^2}{3} - 3(a+b+c) + 8 = \frac{117^2}{3} - 3 \cdot 117 + 8 = 4220,$$
with equality iff $a=b=c=39$. This shows that the minimum number of passwords needed to ensure that at least one mission is networked is $n=2110$. This number is sufficient iff the three missions are of equal size.