(3) Hint: The direction of the torque depends on the direction of the force and the magnitude f the torque depends on the distance of the force from the axis.

Step 1: Find the direction of the torque about the two axes.

(a) Consider the adjacent diagram, where r > r'.
   Torque $\overrightarrow{\mathrm{\tau }}$ about z-axis = r x F which is along $\hat{\mathrm{k}}$

(b) ${\overrightarrow{\mathrm{\tau }}}^{\mathrm{\prime }}={\overrightarrow{\mathrm{r}}}^{\mathrm{\prime }}\times \overrightarrow{\mathrm{F}}\text{ which is along }-\hat{\mathrm{k}}$

Step 2: Find the magnitude of the torque about the two axes.

(c) $|\overrightarrow{\mathrm{\tau }}{|}_{\mathrm{z}}={\mathrm{Fr}}_{\perp }=$ the magnitude of the torque about the z-axis where ${\mathrm{r}}_{\perp }$ is the perpendicular distance between F and z-axis.
$\begin{array}{l}\mathrm{Similarly}, |\overrightarrow{\mathrm{\tau }}{|}_{{\mathrm{z}}^{\mathrm{\prime }}}={\mathrm{Fr}}_{\perp }^{\mathrm{\prime }}\\ \mathrm{Clearly}, {\mathrm{r}}_{\perp }>{\mathrm{r}}_{\perp }^{\mathrm{\prime }}\Rightarrow \left|\overrightarrow{\mathrm{\tau }}{|}_{\mathrm{z}}>\right|\overrightarrow{\mathrm{\tau }}{|}_{{\mathrm{z}}^{\mathrm{\prime }}}\end{array}$

(d) We are always calculating resultant torque about a common axis.

Hence, total torque ${\overrightarrow{\mathrm{\tau }}}_{\mathrm{net}}\ne \overrightarrow{\mathrm{\tau }}+{\overrightarrow{\mathrm{\tau }}}^{\mathrm{\prime }}$, because $\overrightarrow{\mathrm{\tau }}$ and $\overrightarrow{\mathrm{\tau }}$' are not about a common axis.