# A force $\vec{F}=(\hat{i}+2\hat{j}+3\hat{k})N$ acts at a point $(4\hat{i}+3\hat{j}-\hat{k})m$. Then the magnitude of torque about the point $(\hat{i}+2\hat{j}+\hat{k})m$ will be $\sqrt{x}\; N-m$. The value of $x$ is ________. Option: 1 195 Option: 2 200 Option: 3 390 Option: 4 410

$\begin{array}{l} \vec{\tau}=\left(\overrightarrow{\mathrm{r}}_{2}-\overrightarrow{\mathrm{r}}_{1}\right) \times \overrightarrow{\mathrm{F}} \\ =[(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})-(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})] \times \overrightarrow{\mathrm{F}} \\ =(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ \tau=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 1 & -2 \\ 1 & 2 & 3 \end{array}\right| \\ =7 \hat{\mathrm{i}}-11 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\ |\vec{\tau}|=\sqrt{195} \end{array}$

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