Moment about the point $\overrightarrow i + 2\overrightarrow j - \overrightarrow k $of a force represented by $\overrightarrow i + 2\overrightarrow j + \overrightarrow k $ acting through the point $2\overrightarrow i + 3\overrightarrow j + \overrightarrow k $ is
(A) $3\overrightarrow i + \overrightarrow j - \overrightarrow k $
(B) $3\overrightarrow i - \overrightarrow j + \overrightarrow k $
(C) $ - 3\overrightarrow i + \overrightarrow j + \overrightarrow k $
(D) $3\overrightarrow i + \overrightarrow j + \overrightarrow k $

Hint: When a force is applied on a point lying on a line passing through another point then the force produces a torque on another point which lies on the line passing through the point of application of force. This torque is also known as the moment.
Formula used
$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F $
$\overrightarrow \tau $ is the torque(moment of force)s, $\overrightarrow r $is distance vector, $\overrightarrow F $ is the force vector.

Complete Step-by-step solution
$\overrightarrow r $is the distance vector which is the difference of position vector of the points which represent the point of application of force and the point of reference.
$ \Rightarrow \overrightarrow r = (2\overrightarrow i + 3\overrightarrow j + \overrightarrow k ) - (\overrightarrow i + 2\overrightarrow j - \overrightarrow k )$
$ \Rightarrow \overrightarrow r = \overrightarrow i + \overrightarrow j + 2\overrightarrow k $
Given,
$ \Rightarrow \overrightarrow F = \overrightarrow i + 2\overrightarrow j + \overrightarrow k $
$\overrightarrow F $ is the force vector.
We know that,
$ \Rightarrow \overrightarrow \tau = \overrightarrow r \times \overrightarrow F $
$\overrightarrow \tau $ is the torque vector.
\[ \Rightarrow \overrightarrow \tau = (\overrightarrow i + \overrightarrow j + 2\overrightarrow k ) \times (\overrightarrow i + 2\overrightarrow j + \overrightarrow k )\]
$ \Rightarrow \overrightarrow \tau = - 3\overrightarrow i + \overrightarrow j + \overrightarrow k $

Hence the correct answer is (C) $ - 3\overrightarrow i + \overrightarrow j + \overrightarrow k $.

Additional information
For a force to produce torque(moment of force) then force must not be parallel to the distance vector and should not be parallel to the axis about which the body is going to rotate.
The ease with which a body will rotate after applying torque(moment of force) is called the moment of inertia of that body along a given axis.
In the above calculation, we wrote $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F $ which means that the torque(moment of force) vector is the vector obtained through the cross product/vector product of $\overrightarrow r $ and $\overrightarrow F $.
$\overrightarrow \tau $ is the counterpart of $\overrightarrow F $ i.e. the significance of $\overrightarrow F $ in translation motion is similar to that of $\overrightarrow \tau $in rotational motion.

Note:
Torque(moment of force) is the amount of force that can cause an object to rotate about an axis. Just as force is what causes an object to accelerate in linear kinematics, torque causes an object to gain an angular acceleration. Torque is a vector quantity.