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Let $\bar{a}=3 \hat{i}+2 \hat{j}+x \hat{k}$ and $\bar{b}=\hat{i}-\hat{j}+\widehat{k}$, for some real $x$. Then $|\bar{a} \times \bar{b}|=r$ is possible, if
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$r \geq 5 \sqrt{\frac{3}{2}}$
$\begin{aligned} & \vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & x \\ 1 & -1 & 1\end{array}\right|=(2+x) \hat{i}+(x-3) \hat{j}+(-3-2) \hat{k} \\ & \text { now } r=|\vec{a} \times \vec{b}|=\sqrt{(2+x)^2+(x-3)^2+(-5)^2} \\ & =\sqrt{2 x^2-2 x+38} \\ & =\sqrt{2\left(x-\frac{1}{2}\right)^2+\frac{75}{2}} \\ & \Rightarrow r_{\min }=\sqrt{\frac{75}{2}} \text { at } x=\frac{1}{2} \\ & \Rightarrow r \geq 5 \cdot \sqrt{\frac{3}{2}}\end{aligned}$
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