The sum of the common terms of the following three arithmetic progressions.
$$3, 7, 11, 15, \ldots, 399$$
$$2, 5, 8, 11, \ldots, 359$$ and
$$2, 7, 12, 17, \ldots, 197$$, is equal to ______.

We need to find the sum of common terms of three arithmetic progressions:

AP1: $$3, 7, 11, 15, \ldots, 399$$ (first term $$a_1 = 3$$, common difference $$d_1 = 4$$)

AP2: $$2, 5, 8, 11, \ldots, 359$$ (first term $$a_2 = 2$$, common difference $$d_2 = 3$$)

AP3: $$2, 7, 12, 17, \ldots, 197$$ (first term $$a_3 = 2$$, common difference $$d_3 = 5$$)

A common term $$a$$ must satisfy all three conditions simultaneously: $$a \equiv 3 \pmod{4}$$, $$a \equiv 2 \pmod{3}$$, $$a \equiv 2 \pmod{5}$$.

Using the Chinese Remainder Theorem with $$\text{lcm}(4, 3, 5) = 60$$, from $$a \equiv 2 \pmod{5}$$ we write $$a = 5k + 2$$. Then imposing $$a \equiv 3 \pmod{4}$$ gives $$5k + 2 \equiv 3 \pmod{4} \Rightarrow 5k \equiv 1 \pmod{4} \Rightarrow k \equiv 1 \pmod{4}$$, so $$k = 4m + 1$$ and $$a = 5(4m + 1) + 2 = 20m + 7$$.

Next, imposing $$a \equiv 2 \pmod{3}$$ leads to $$20m + 7 \equiv 2 \pmod{3} \Rightarrow 2m + 1 \equiv 2 \pmod{3} \Rightarrow m \equiv 2 \pmod{3}$$, hence $$m = 3j + 2$$ and $$a = 20(3j + 2) + 7 = 60j + 47$$.

The common terms therefore form the sequence $$47, 107, 167, 227, 287, 347, \ldots$$.

Applying the bounds from each AP, the most restrictive upper bound is from AP3: $$a \leq 197$$, so the valid common terms are $$47, 107, 167$$ since $$227 > 197$$.

Verification shows that for $$47$$, $$(47 - 3)/4 = 11$$ ✓, $$(47 - 2)/3 = 15$$ ✓, and $$(47 - 2)/5 = 9$$ ✓.

For $$107$$, $$(107 - 3)/4 = 26$$ ✓, $$(107 - 2)/3 = 35$$ ✓, and $$(107 - 2)/5 = 21$$ ✓.

For $$167$$, $$(167 - 3)/4 = 41$$ ✓, $$(167 - 2)/3 = 55$$ ✓, and $$(167 - 2)/5 = 33$$ ✓.

The sum of these common terms is 321.