A force  $\overrightarrow{F}=4\widehat{i}+3\widehat{j}+4\widehat{k}$ N  is applied on an intersection point of x = 2 m  plane and X – axis. The magnitude of torque of this force about a point (2 m, 3 m, 4 m) is _____Nm.

given  $\text{ force, }\mathrm{F}=4\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+4\widehat{\mathrm{k}}$We know that$\text{ torque, }\mathrm{\tau }=\mathrm{r}\times \mathrm{F}$Where , r is the perpendicular            

         distance  
$\begin{array}{l}\mathrm{r}=\left(2\widehat{\mathrm{i}}\right)-(2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+4\widehat{\mathrm{k}})=-3\widehat{\mathrm{j}}-4\widehat{\mathrm{k}}\\ \Rightarrow \mathrm{\tau }=\mathrm{r}\times \mathrm{F}=\left|\begin{array}{ccr}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 0& -3& -4\\ 4& 3& 4\end{array}\right|\\ \Rightarrow \mathrm{\tau }=\widehat{\mathrm{i}}(-12+12)-\widehat{\mathrm{j}}(0+16)+\widehat{\mathrm{k}}(0+12)\\ \Rightarrow \mathrm{\tau }=-16\widehat{\mathrm{i}}+12\widehat{\mathrm{k}}\end{array}$
Magnitude of torque 
$\begin{array}{l}\left|\mathrm{\tau }\right|=\sqrt{(16{)}^2+(12{)}^2}\\ =\sqrt{256+144}=\sqrt{400}\\ \left|\mathrm{\tau }\right|=20\end{array}$