A slab is subjected to two forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ of same magnitude $F$ as shown in the figure. Force $\overrightarrow{F_2}$ is in $XY$ - plane while force $\overrightarrow{F_1}$ acts along $Z$- axis at the point $(2\hat{i}+3\hat{j})$ The moment of these forces about point $O$ will be

The correct option is A $(3\hat{i}-2\hat{j}+3\hat{k})F$
Calculate the torque for $\overrightarrow{F_1}$ force.

From figure
${\overrightarrow{F}}_1=F\hat{k}$

$\overrightarrow{r}=2\hat{i}+3\hat{j}$
Torque for $\overrightarrow{F_1}$ force
${\overrightarrow{\tau }}_{F_1}={\overrightarrow{r}}_1\times {\overrightarrow{F}}_1$

$=3F\hat{i}+2F(-\hat{j})$

Calculate the torque for $\overrightarrow{F_2}$ force.
From figure
${\overrightarrow{r}}_2=0\hat{i}+6\hat{j}$

${\overrightarrow{F}}_2=F_2\sin {30}^{\circ }(-\hat{i})+F_2\cos {30}^{\circ }(-\hat{j})$

${\overrightarrow{F}}_2=\frac{F}{2}(-\hat{i})+\frac{F\sqrt{3}}{2}(-\hat{j})$

Torque for $\overrightarrow{F_2}$ force
${\overrightarrow{\tau }}_{F_2}={\overrightarrow{r}}_2\times {\overrightarrow{F}}_2=3F\hat{k}$

Calculate the net torque.

Formula used: ${\overrightarrow{\tau }}_{net}={\overrightarrow{\tau }}_{F_1}+{\overrightarrow{\tau }}_{F_2}$

${\overrightarrow{\tau }}_{net}={\overrightarrow{\tau }}_{F_1}+{\overrightarrow{\tau }}_{F_2}$

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

${\overrightarrow{\tau }}_{net}=(3\hat{i}-2\hat{j}+3\hat{k})F$

Final answer :$\left(d\right)$