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

Let $$\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$$, $$\vec{b}=\hat{i}+\hat{j}-\hat{k}$$ and $$\vec{c}=2\hat{i}+2\hat{j}-\hat{k}$$.

The required unit vector $$\vec{u}$$ must satisfy two conditions:
(i) $$\vec{u}\cdot\vec{a}=0$$  (perpendicular to $$\vec{a}$$)
(ii) $$\vec{u}$$ lies in the plane of $$\vec{b}$$ and $$\vec{c}$$, i.e. $$\vec{u}\cdot\vec{n}=0$$ where $$\vec{n}=\vec{b}\times\vec{c}$$ is the normal to that plane.

Step 1: Find the normal to the plane of $$\vec{b},\vec{c}$$.
$$\vec{n}=\vec{b}\times\vec{c} =\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\[2pt] 1&1&-1\\[2pt] 2&2&-1 \end{vmatrix} =1\hat{i}-1\hat{j}+0\hat{k} =\hat{i}-\hat{j}$$

Step 2: Produce a vector that is simultaneously perpendicular to $$\vec{a}$$ and to $$\vec{n}$$. The cross-product $$\vec{a}\times\vec{n}$$ achieves this because it is orthogonal to both of its factors.

Compute $$\vec{a}\times\vec{n}$$:
$$ \vec{a}\times\vec{n} =\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\[2pt] 2&-1&2\\[2pt] 1&-1&0 \end{vmatrix} =2\hat{i}+2\hat{j}-\hat{k} $$

Check the two required perpendicularities:
$$\bigl(2\hat{i}+2\hat{j}-\hat{k}\bigr)\cdot\bigl(2\hat{i}-\hat{j}+2\hat{k}\bigr)=4-2-2=0$$ $$\bigl(2\hat{i}+2\hat{j}-\hat{k}\bigr)\cdot(\hat{i}-\hat{j})=2-2=0$$ Hence the vector $$\vec{v}=2\hat{i}+2\hat{j}-\hat{k}$$ meets both conditions.

Step 3: Convert $$\vec{v}$$ to a unit vector.
Magnitude: $$\lVert\vec{v}\rVert=\sqrt{2^2+2^2+(-1)^2}=3$$
Unit vector: $$\vec{u}=\dfrac{\vec{v}}{\lVert\vec{v}\rVert}=\dfrac{2\hat{i}+2\hat{j}-\hat{k}}{3}$$

Therefore the required unit vector is $$\displaystyle\frac{2\hat{i}+2\hat{j}-\hat{k}}{3}$$.

Option D which is: $$\dfrac{2\hat{i} + 2\hat{j} - \hat{k}}{3}$$