To determine the moment of the forces about point O, we need to calculate the moments of each force and then combine them vectorially.

First, consider force $$\overrightarrow{F_1}$$. This force acts along the z-axis at the point $$(2\overrightarrow{i} + 3\overrightarrow{j})$$.

The position vector of the point of application of $$\overrightarrow{F_1}$$ relative to the origin O is:

$$\overrightarrow{r_1} = 2\overrightarrow{i} + 3\overrightarrow{j}$$

Since $$\overrightarrow{F_1}$$ acts along the z-axis, we can express it as:

$$\overrightarrow{F_1} = F\overrightarrow{k}$$

The moment of force $$\overrightarrow{F_1}$$ about point O is given by the cross product:

$$\overrightarrow{M_1} = \overrightarrow{r_1} \times \overrightarrow{F_1}$$

Substitute the values:

$$\overrightarrow{M_1} = (2\overrightarrow{i} + 3\overrightarrow{j}) \times F\overrightarrow{k}$$

Compute the cross product:

$$\overrightarrow{M_1} = 2F(\overrightarrow{i} \times \overrightarrow{k}) + 3F(\overrightarrow{j} \times \overrightarrow{k})$$ $$\overrightarrow{M_1} = 2F(-\overrightarrow{j}) + 3F\overrightarrow{i}$$ $$\overrightarrow{M_1} = -2F\overrightarrow{j} + 3F\overrightarrow{i}$$

So, the moment due to force $$\overrightarrow{F_1}$$ is:

$$\overrightarrow{M_1} = 3F\overrightarrow{i} - 2F\overrightarrow{j}$$

Torque for F₂ force

$${\overrightarrow F _2}$$ = $${F \over 2}\left( { - \widehat i} \right) + {{F\sqrt 3 } \over 2}\left( { - \widehat j} \right)$$

$${\overrightarrow r _1} = 0\widehat i + 6\widehat j$$

$${\overrightarrow \tau _{_{{F_2}}}} = \overrightarrow {{r_1}} \times \overrightarrow {{F_1}} = 3F\widehat k$$

$${\overrightarrow \tau _{net}} = {\overrightarrow \tau _{{F_1}}} + {\overrightarrow \tau _{{f_2}}}$$

$$ = 3F\widehat i + 2F\left( { - \widehat j} \right) + 3F\left( {\widehat k} \right)$$