Let $A$, $B$, $C$ be committees and let them have exactly $a$, $b$, $c$ persons, respectively. To each person assign a point in the plane so that no three points are collinear. We connect the points corresponding to persons that know each other by a blue segment, and for those that do not know each other by a red segment. Points corresponding to persons in the same committee are not connected.

First, let us count the monochromatic triangles (all three sides are of the same colour). For each pair of persons $x \in A$, $y \in B$ there are exactly $10$ monochromatic triangles (if $xy$ is a blue segment, there are $10$ points $z \in C$ such that segments $xz$ and $yz$ are blue, and analogously for red segments). Hence, the number of monochromatic triangles is $10ab$. In the same way we conclude that the number of monochromatic triangles is $10bc$ and $10ca$, so we conclude $a = b = c$.

The number of all triangles is $a^3$, while the number of monochromatic triangles is $10a^2$. Let us count bichromatic triangles. For each pair $x \in A$, $y \in B$ we have exactly $10$ bichromatic triangles such that the segments $xz$ and $yz$ have the same colour, i.e. there are $10a^2$ such triangles, while there are $10a^2$ triangles such that $yx$ and $zx$ are of the same colour, and $10a^2$ triangles such that $xy$ and $zy$ are of the same colour. Hence there are $30a^2$ bichromatic triangles.

Finally, we have $a^3 = 10a^2 + 30a^2$, so $a = 40$. This means that there are in total $120$ members in all three committees.