The moment of force \( \overrightarrow{\mathrm{F}} \) about the point \( (3,5) \) is \( 6 \hat{\mathrm{k}} \) and its moment about the point \( (1,-1) \) is \( -6 \hat{\mathrm{k}} \), then its moment about the point \( \cdots \cdots \cdots=\overrightarrow{0} \) \( \begin{array}{llll}\text { (a) }(-1,-3) & \text { (b) }(2,2) & \text { (c) }(2,6) & \text { (d) }(1,3)\end{array} \)

To solve this, we can use the relationship between the moments at different points. The moment of a force about any point can be derived from its moment about another point and the position vector difference. Given that the moment of force about the points \( (3, 5) \) is \( 6 \hat{k} \) and about \( (1, -1) \) is \( -6 \hat{k} \), we can express the moments with respect to point \( (x, y) \) as: \[ M_{(x, y)} = M_{(3, 5)} + \overrightarrow{r_{(3, 5) \to (x, y)}} \times \overrightarrow{F} \] Now, the change in moment can be calculated: \[ M_{(3, 5)} - M_{(1, -1)} = 6 \hat{k} - (-6 \hat{k}) = 12 \hat{k} \] This implies that the difference in position vector times the force is equal to \( 12 \hat{k} \). We also need to find the point where the moment becomes zero. Let’s calculate the moment in relation to the unknown point \( (x, y) \): If we test the options, we notice that the distance vector from \( (1, -1) \) to our candidate points should keep the relationship valid. After checking the options, option (d) \( (1, 3) \) gives us: \[ \overrightarrow{r} = (1 - 1, 3 - (-1)) = (0, 4) \] This would result in a zero moment since it sums up to the initial setting. Thus, the answer is \( (1, 3) \). The logic behind the problem employs vector mathematics and the principle of moments in physics. It's a great example of how understanding spatial relationships can provide insights into force dynamics!