Here's what the textbook says
The equilibrium condition for the torques is true for any choice of the axis about which the torques are calculated. To prove this statement, we consider a rigid body on which many forces act. Relative to the origin O, force $\overrightarrow{F_1}$ is applied at the point located at $\overrightarrow{r_1}$, force $\overrightarrow{F_2}$ at $\overrightarrow{r_2}$ and so on. The net torque about an axis through O is therefore
$\begin{aligned} \overrightarrow{\boldsymbol{\tau}}_{O} &=\overrightarrow{\boldsymbol{\tau}}_{1}+\overrightarrow{\boldsymbol{\tau}}_{2}+\cdots+\overrightarrow{\boldsymbol{\tau}}_{N} \\ &=\overrightarrow{\mathbf{r}}_{1} \times \overrightarrow{\mathbf{F}}_{1}+\overrightarrow{\mathbf{r}}_{2} \times \overrightarrow{\mathbf{F}}_{2}+\cdots+\overrightarrow{\mathbf{r}}_{N} \times \overrightarrow{\mathbf{F}}_{N} \end{aligned}$
Suppose a point P is located at displacement with respect to O. The point of application of $\overrightarrow{F_1}$ with respect to P, is$(\overrightarrow{r_1} - \overrightarrow{r_P})$. The torque about P is
$\begin{aligned} \overrightarrow{\boldsymbol{\tau}}_{P}=&\left(\overrightarrow{\mathbf{r}}_{1}-\overrightarrow{\mathbf{r}}_{P}\right) \times \overrightarrow{\mathbf{F}}_{1}+\left(\overrightarrow{\mathbf{r}}_{2}-\overrightarrow{\mathbf{r}}_{P}\right) \times \overrightarrow{\mathbf{F}}_{2} \\ &+\cdots+\left(\overrightarrow{\mathbf{r}}_{N}-\overrightarrow{\mathbf{r}}_{P}\right) \times \overrightarrow{\mathbf{F}}_{N} \\=&\left[\overrightarrow{\mathbf{r}}_{1} \times \overrightarrow{\mathbf{F}}_{1}+\overrightarrow{\mathbf{r}}_{2} \times \overrightarrow{\mathbf{F}}_{2}+\cdots+\overrightarrow{\mathbf{r}}_{N} \times \overrightarrow{\mathbf{F}}_{N}\right] \\ &-\left[\overrightarrow{\mathbf{r}}_{P} \times \overrightarrow{\mathbf{F}}_{1}+\overrightarrow{\mathbf{r}}_{P} \times \overrightarrow{\mathbf{F}}_{2}+\cdots+\overrightarrow{\mathbf{r}}_{P} \times \overline{\mathbf{F}}_{N}\right] \end{aligned}$
The first group of terms in the brackets gives $\tau_O$. We can rewrite the second group by removing the constant factor of $\overrightarrow{r_P}$
$\begin{aligned} \overrightarrow{\boldsymbol{\tau}}_{P} &=\overrightarrow{\boldsymbol{\tau}}_{O}-\left[\overrightarrow{\mathbf{r}}_{P} \times\left(\overrightarrow{\mathbf{F}}_{1}+\overrightarrow{\mathbf{F}}_{2}+\cdots+\overrightarrow{\mathbf{F}}_{N}\right)\right] \\ &=\overrightarrow{\boldsymbol{\tau}}_{O}-\left[\overrightarrow{\mathbf{r}}_{P} \times\left(\sum \overrightarrow{\mathbf{F}}_{\mathrm{ext}}\right)\right] \\ &=\overrightarrow{\boldsymbol{\tau}}_{O} \end{aligned}$
where we make the last step because $\sum \overrightarrow{F_{ext}}=0$ for a body in translational equilibrium. Thus the torque about any two points has the same value when the body is in translational equilibrium.
What is the physical meaning of this? How do we apply this in questions? An example too will help, because I'm having trouble visualizing this.
(source: Physics by Halliday, Resnick, Krane; 5th edition; Pg 188, Rotational Dynamics)