If $\overrightarrow{\mathrm{A}}=4\mathrm{N},\overrightarrow{\mathrm{B}}=3\mathrm{N}$ the value of ${\left|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right|}^2+{\left|\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}\right|}^2$ then

Given,

$$\begin{gathered} \overrightarrow{\mathrm{A}}=4\mathrm{N} \\ \overrightarrow{\mathrm{B}}=3\mathrm{N} \end{gathered}$$

Here,

$\left|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right|=\left|\mathrm{A}\right|\left|\mathrm{B}\right|\mathrm{Sin\theta }$

$\left|\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}\right|=\left|\mathrm{A}\right|\left|\mathrm{B}\right|\mathrm{Cos\theta }$

${\left|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right|}^2+{\left|\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}\right|}^2={\left(\left|\mathrm{A}\right|\left|\mathrm{B}\right|\right)}^2\left({\mathrm{Sin\theta }}^2+{\mathrm{Cos\theta }}^2\right)$

${\left|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right|}^2+{\left|\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}\right|}^2={\left(\left|\mathrm{A}\right|\left|\mathrm{B}\right|\right)}^2=16\times 9=144\mathrm{N}$

Hence the correct answer is 144N.