1. We need the common terms of the three infinite arithmetic progressions:

$AP(1;3)=\{1,4,7,10,\dots\}$ $AP(2;5)=\{2,7,12,17,\dots\}$ $AP(3;7)=\{3,10,17,24,\dots\}$

If their intersection is again an arithmetic progression $AP(a;d)$, then its terms must satisfy all three congruences.

2. Let a common term be $x$. Then

$x \equiv 1 \pmod{3}$ $x \equiv 2 \pmod{5}$ $x \equiv 3 \pmod{7}$

So we solve this system of congruences.

3. First combine the first two:

$x \equiv 1 \pmod{3}$ $x \equiv 2 \pmod{5}$

Numbers congruent to $1 \pmod 3$ are:

$1,4,7,10,13,16,19,22,25,28,\dots$

Among these, those congruent to $2 \pmod 5$ are:

$7,22,37,52,\dots$

Hence

$x \equiv 7 \pmod{15}$

4. Now combine with

$x \equiv 3 \pmod{7}$

Write

$x=7+15k$

Substitute into the third congruence:

$7+15k \equiv 3 \pmod{7}$

Since $7 \equiv 0 \pmod 7$ and $15 \equiv 1 \pmod 7$,

$k \equiv 3 \pmod 7$

So

$k=3+7t$

Thus

$x=7+15(3+7t)=7+45+105t=52+105t$

Therefore,

$AP(1;3) \cap AP(2;5) \cap AP(3;7)=AP(52;105)$

So

$a=52,\quad d=105$

5. Hence

$a+d=52+105=157$

So the required integer is

$\boxed{157}$