The ages are:
7, 12, 21
In order to see this,
let $a$ and $b$ be the ages of your daughters, and let $c$ be the age of your son. We now that $a$ and $c$ have a common divisor, from this it follows that they also have a common prime divisor, let this prime divisor be $p$. Similarly, $b$ and $c$ have a common prime divisor $q$. We assume that $p \leq q$.
Because $a$ and $b$ are coprime, we have that $p$ does not divide $b$, from this it also follows that $p$ does not divide $a + b + c = 40$. Similarly, $q$ does not divide 40 either.
It also follows that $pq$ is a divisor of $c$, so $c \geq pq$, similarly $b \geq q$. We also have that $p \geq 3$ because $p$ is a prime that does not divide 40. This means that $40 = a + b + c \geq q + pq \geq 4q$. This means that $q \leq 10$.
So $p$ and $q$ are two primes not dividing 40 with $p < q \leq 10$. This is only possible if $p = 3$ and $q = 7$. So $c$ is a multiple of $21$, and therefore $c = 21$. It follows that $a + b = 19$, and $a$ is divisible by 3 and $b$ is divisible by $7$. This is only possible if $a = 12$ and $b = 7$. So this are the only possible ages of the children.
The above implicitly assumes that all ages are non-zero. If we drop this assumption, we get some more solutions (Although I do not think this is intended):
3, 37, 0
7, 33, 0
9, 31, 0
11, 29, 0
13, 27, 0
17, 23, 0
19, 21, 0