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(D) Let $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ be the required unit vector.Since $\vec{v}$ is perpendicular to $\vec{a} = 2\hat{i} - \hat{j} + 2\h

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$\frac{2\hat{i} + 2\hat{j} - \hat{k}}{3}$

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(D) Let $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ be the required unit vector.Since $\vec{v}$ is perpendicular to $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$,we have $\vec{v} \cdot \vec{a} = 0$.$2x - y + 2z = 0$ ...... $(i)$Since $\vec{v}$ is coplanar with $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} + 2\hat{j} - \hat{k}$,$\vec{v}$ can be written as a linear combination of $\vec{b}$ and $\vec{c}$.$\vec{v} = p(\hat{i} + \hat{j} - \hat{k}) + q(2\hat{i} + 2\hat{j} - \hat{k}) = (p + 2q)\hat{i} + (p + 2q)\hat{j} - (p + q)\hat{k}$.Comparing components,$x = p + 2q$,$y = p + 2q$,$z = -(p + q)$.Substitute into $(i)$: $2(p + 2q) - (p + 2q) + 2(-p - q) = 0$.$2p + 4q - p - 2q - 2p - 2q = 0 \Rightarrow -p = 0 \Rightarrow p = 0$.Thus,$x = 2q, y = 2q, z = -q$.Since $\vec{v}$ is a unit vector,$x^2 + y^2 + z^2 = 1$.$(2q)^2 + (2q)^2 + (-q)^2 = 1 \Rightarrow 4q^2 + 4q^2 + q^2 = 1 \Rightarrow 9q^2 = 1 \Rightarrow q = \pm \frac{1}{3}$.For $q = \frac{1}{3}$,$\vec{v} = \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} = \frac{2\hat{i} + 2\hat{j} - \hat{k}}{3}$.This matches option $D$.

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

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