A question that is not repeated

Let's assume that the present ages of Andrew, David and Helen are $a, d$ and $h$ respectively. So,

$a + d+h = 49 ~~~ ........ (1)$

$d = 3h ~~~~ ........(2)$

$a + 5 = 3(d + 5) ~~~ ......(3).$

Now substituting $d = 3h$ in $(1):$

$a + 3h + h = 49 \implies a = 49 - 4h.$

Substituting $a = 49 - 4h$ and $d = 3h$ in $(3):$

$\begin{aligned} 49 - 4h + 5 & = 3(3h + 5) \\ \Rightarrow 54 & = 9h + 15 + 4h \\ \Rightarrow 13h & = 39 \\ \implies h & = 3 \\ \end{aligned}$

Substituting $h = 3$ in $(2):$

$d = 3 \times 3 \implies d = 9.$

Finally, substituting $d = 9$ and $h = 3$ in $(1):$

$a + 9 + 3 = 49 \implies a = 37$

Hence $adh = 37 \times 9 \times 3 = \boxed{999}$