Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the total spin angular momentum $ S $ for a particle with $ s = 1 $ is given by
$$S=\sqrt{s(s+1)}\;\;\hbar$$
and we are given that $s=1$, so
$$S=\sqrt{1(1+1)}\;\;\hbar=\sqrt{2}\;\;\hbar$$
Plug the known;
$$S =\sqrt{2}\;(1.05 \times 10^{-34} )$$
$$S= \color{red}{\bf 1.48 \times 10^{-34}} \;\rm J\cdot s$$
$$\color{blue}{\bf [b]}$$
The spin quantum number $ m_s $ can take three possible values
$$ m_s = -1, 0, 1 $$
These represent the possible projections of the spin angular momentum along the z-axis.
$$\color{blue}{\bf [c]}$$
The diagram illustrates the three possible orientations of the spin vector $ \vec{S} $ relative to the z-axis. The values of $ S_z $ (the projection of $ \vec{S} $ on the z-axis) are $ \hbar $, 0, and $ -\hbar $, corresponding to $ m_s = 1 $, $ m_s = 0 $, and $ m_s = -1 $, respectively. These vectors form a circular arrangement where the magnitude of $ \vec{S} $ remains constant at $ \sqrt{2}\hbar $, but the projection changes according to the allowed values of $ m_s $.
This figure visually shows how the spin angular momentum can point in different directions, while the overall spin magnitude stays the same, constrained by quantum mechanics to specific orientations based on $ m_s $.
