If the resultant force on a body is zero, why is it so that the total torque about any line perpendicular to the plane of forces is equal ?


The statement that you have made is true in general.
This is because if there is no resultant the only thing have left is a couple.
A couple is two equal, opposite and parallel (non co-linear) forces and has the property that its torque is independent of position about which the torque is evaluated.

In simple terms I can show you this in terms of three coplanar forces $\vec F_1, \vec F_2$ and $\vec F_3$ acting at points $A, B$ and $C$ as in the diagram below.

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The resultant of these three forces is zero, $\vec F_1 + \vec F_2 + \vec F_3 = 0$

Just for ease of drawing set up some axes such that force $\vec F_1$ only have an x-component $F_{1 \rm x}$.

Noting that $F_{2\rm x} = F_{1\rm x} + F_{3\rm x}$ and $F_{2\rm y} =F_{3\rm y}$ there are three couples acting on this body with fixed magnitudes $F_{1\rm x}y_{12}, F_{3\rm x}y_{23}$ and $F_{3\rm y}x_{23}$.

With extra hand-waving you can extend this to any number of forces and into three dimensions.

Alternatively . .

Suppose that you have $n$ forces $\vec F_n$ acting on a body and the resultant force is zero then $\sum \limits_n \vec F_n =0$.

The torque about any point is $\sum \limits_n \vec r_n \times \vec F_n $ where $\vec r_n$ are the position vectors of the point of application of the forces.

Now move from the origin by $\vec r$ and the torque about that new point is

$\sum \limits_n (\vec r_n - \vec r) \times \vec F_n = \sum \limits_n \vec r_n \times \vec F_n - \vec r \times \sum \limits_n \vec F_n = \sum \limits_n \vec r_n \times \vec F_n - \vec r \times \vec 0 = \sum \limits_n \vec r_n \times \vec F_n$ as before.


I think there is a misnomer here. A pure torque (if net forces are zero) does not have a location associated with it. It is shared identically throughout the rigid body.

Only a torque (force moment) as a result of a force at a distance changes value with location (due to the distance to the force changing).

Torque transforms from point A to point B with the law $$\vec{\tau}_B = \vec{\tau}_A + \vec{r}_{A/B} \times \vec{F} $$

If the force $\vec{F}$ is zero then the torque is identical on all the points ($\vec{\tau}_B = \vec{\tau}_A$).

PS. The transformation law is identical to angular momentum $$\vec{L}_B = \vec{L}_A + \vec{r}_{A/B} \times \vec{p} $$ and also with velocities $$\vec{v}_B = \vec{v}_A + \vec{r}_{A/B} \times \vec{\omega} $$


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