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(A) Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.First,we find the cross product $\vec{a} 	imes \vec{b}

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$\frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k})$

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(A) Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.First,we find the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 1 & 2 & -1 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(-1 - 3) + \hat{k}(2 - (-2)) = -4\hat{i} + 4\hat{j} + 4\hat{k}$.The magnitude of the cross product is $|\vec{a} 	imes \vec{b}| = \sqrt{(-4)^2 + 4^2 + 4^2} = \sqrt{16 + 16 + 16} = \sqrt{48} = 4\sqrt{3}$.The unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by $\pm \frac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|} = \pm \frac{-4\hat{i} + 4\hat{j} + 4\hat{k}}{4\sqrt{3}} = \pm \frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.Comparing this with the given options,the correct option is $\frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k})$.

The unit vector perpendicular to both the vectors $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ is

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