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

The correct option is A $(3\hat{i}-2\hat{j}+3\hat{k})F$
From the given figure,

$\overrightarrow{F_1}=F\hat{k}$
$\overrightarrow{F_2}=F\sin {30}^{\circ }(-\hat{i})+F\cos {30}^{\circ }(-\hat{j})=\frac{F}{2}(-\hat{i})+\frac{F\sqrt{3}}{2}(-\hat{j})$
$\overrightarrow{r_1}=2\hat{i}+3\hat{j}$
$\overrightarrow{r_2}=0\hat{i}+6\hat{j}$

Torque due to force $F_1$,

$\overrightarrow{{\tau }_{F_1}}=\overrightarrow{r_1}\times \overrightarrow{F_1}$

$\Rightarrow \overrightarrow{{\tau }_{F_1}}=(2\hat{i}+3\hat{j})\times F\hat{k}$

$\Rightarrow \overrightarrow{{\tau }_{F_1}}=3F\hat{i}+2F(-\hat{j})$

Torque due to force $F_2$,

$\overrightarrow{{\tau }_{F_2}}=\overrightarrow{r_2}\times \overrightarrow{F_2}$

$\Rightarrow \overrightarrow{{\tau }_{F_2}}=6\hat{j}\times \left(\frac{F}{2}\right(-\hat{i})+\frac{F\sqrt{3}}{2}(-\hat{j}\left)\right)$

$\Rightarrow \overrightarrow{{\tau }_{F_2}}=3F\left(\hat{k}\right)$

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

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

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

Hence, option $\left(A\right)$ is correct.