$\vec{\tau}=\vec{r} \times \vec{F}$
To, we can find Torque by finding the cross product of $\vec{r}$ and $\vec{F}$
We have:
$\vec{r}=3 \hat{\imath}+2 \hat{\jmath}+3 \hat{k} m$
$\vec{F}=2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k} N$
So, torque will be:
$\vec{\tau}=\left|\begin{array}{ccc}{\hat{\imath}} & {\hat{\jmath}} & {\hat{k}} \\ {3} & {2} & {3} \\ {2} & {-3} & {4}\end{array}\right|$
$\Longrightarrow \vec{\tau}=\hat{\imath}((2)(4)-(-3)(3))-\hat{\jmath}((3)(4)-(2)(3))+\hat{k}((3)(-3)-(2)(2))$
$\Longrightarrow \vec{\tau}=\hat{\imath}(8+9)-\hat{\jmath}(12-6)+\hat{k}(-9-4)$
$\Longrightarrow[\vec{\tau}=17 \hat{\imath}-6 \hat{\jmath}-13 \hat{k}, N m]$
This is the torque of the force acting about Origin.
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Statement $(I)$ : Dimensions of specific heat is $\left[\mathrm{L}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]$
Statement $(II)$ : Dimensions of gas constant is $\left[\mathrm{ML}^2 \mathrm{~T}^{-1} \mathrm{~K}^{-1}\right]$