Rank and nullity of a 3D rotation matrix

Let $R$ be the $3\times3$ matrix obtained by composing counter-clockwise rotations through $\alpha$ radians about the $x$-axis, $\beta$ radians about the $y$-axis and $\gamma$ radians about the $z$-axis. Whatever the angles, what is $(\operatorname{rank}R)^2+(\operatorname{null}R)^2$?

$9$. Rotations preserve length, so only the zero vector maps to zero: the nullity is $0$ and the rank is $3$, giving $9+0=9$.