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(D) Let $\vec{a}$ be the required unit vector. Since $\vec{a}$ is coplanar with $\vec{c} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{d} = 2 \hat{i}-\hat{j}-\

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

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(D) Let $\vec{a}$ be the required unit vector. Since $\vec{a}$ is coplanar with $\vec{c} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{d} = 2 \hat{i}-\hat{j}-\hat{k}$, it must be perpendicular to the vector $\vec{n} = \vec{c} 	imes \vec{d}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & -1 & -1 \end{vmatrix} = \hat{i}(-1+1) - \hat{j}(-1-2) + \hat{k}(-1-2) = 0\hat{i} + 3\hat{j} - 3\hat{k}$.Since $\vec{a}$ is also perpendicular to $\vec{b} = \hat{i}-2 \hat{j}+3 \hat{k}$, $\vec{a}$ must be parallel to $\vec{b} 	imes \vec{n}$.$\vec{b} 	imes \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 0 & 3 & -3 \end{vmatrix} = \hat{i}(6-9) - \hat{j}(-3-0) + \hat{k}(3-0) = -3\hat{i} + 3\hat{j} + 3\hat{k}$.The unit vector is $\pm \frac{\vec{b} 	imes \vec{n}}{|\vec{b} 	imes \vec{n}|} = \pm \frac{-3\hat{i} + 3\hat{j} + 3\hat{k}}{\sqrt{(-3)^2 + 3^2 + 3^2}} = \pm \frac{-3\hat{i} + 3\hat{j} + 3\hat{k}}{\sqrt{27}} = \pm \frac{-3\hat{i} + 3\hat{j} + 3\hat{k}}{3\sqrt{3}} = \pm \frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k})$.Note: The vector $\pm \frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})$ is equivalent to $\pm \frac{1}{\sqrt{3}}(-\hat{i}+\hat{j}+\hat{k})$.

The unit vector perpendicular to the vector $\hat{i}-2 \hat{j}+3 \hat{k}$ and coplanar with the vectors $\hat{i}+\hat{j}+\hat{k}$ and $2 \hat{i}-\hat{j}-\hat{k}$ is

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