1. Write the three APs in general form

• First AP: $3,7,11,15,\dots,399$

  This has first term $3$ and common difference $4$. So its general term is $a_1=3+4k \quad (k\ge 0).$ Hence terms satisfy $x\equiv 3 \pmod 4.$

• Second AP: $2,5,8,11,\dots,359$

  First term $2$, common difference $3$. So $a_2=2+3m \quad (m\ge 0),$ hence $x\equiv 2 \pmod 3.$

• Third AP: $2,7,12,17,\dots,197$

  First term $2$, common difference $5$. So $a_3=2+5n \quad (n\ge 0),$ hence $x\equiv 2 \pmod 5.$

We need numbers common to all three APs.

2. Solve the congruences

A common term $x$ must satisfy: $x\equiv 3 \pmod 4, \quad x\equiv 2 \pmod 3, \quad x\equiv 2 \pmod 5.$

Since $x\equiv 2 \pmod 3 \quad \text{and} \quad x\equiv 2 \pmod 5,$ we get $x\equiv 2 \pmod{15}.$

So let $x=15t+2.$

Now impose $15t+2\equiv 3 \pmod 4.$ Since $15\equiv 3 \pmod 4$, this gives $3t+2\equiv 3 \pmod 4$ $3t\equiv 1 \pmod 4.$

As $3\equiv -1 \pmod 4$, we get $-t\equiv 1 \pmod 4 \implies t\equiv 3 \pmod 4.$

So $t=4r+3.$

Hence $x=15(4r+3)+2=60r+47.$

Therefore, the common terms form an AP: $47,107,167,227,287,347,\dots$ with common difference $60$.

3. Find which of these lie in all the given finite APs

A common term must not exceed the last term of any of the three lists. The smallest last term is $197$ (from the third AP), so we only need common terms $\le 197$.

From $47,107,167,227,\dots$ those not exceeding $197$ are: $47,107,167.$

So the common terms are exactly these three.

4. Find their sum

$47+107+167 = 154+167 = 321.$

Thus, the required sum is $\boxed{321}.$

5. Comparison with stored answer

Stored correct answer = $321$.

Our derived answer is also $321$, so it agrees.