Aged problem!!

let the family members be $A,B.C.D.E$ and $F$ , so

$3+7+C+D+E+F=106$ , the sum of the ages of $C,D,E,F$ is $106-10=96$ .

Five years ago, the youngest (3 year old) was not yet born. So the sum of their ages was

$2+96-(4*5)=2+96-20=$ $\color{#D61F06}\boxed{78}$

$3+7+a+b+c+d = 106$

$\Rightarrow a+b+c+d = 96\dots(1)$

Five years ago :

$0+2+(a-5)+(b-5)+(c-5)+(d-5)$

$\Rightarrow a+b+c+d+2-5(4)$

$\Rightarrow 96+2-20 = \boxed{78}$