A unit vector is coplanar to the vectors $\vec{a}=\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}+\hat{k}$ and perpendicular to the vector $\vec{c}=\hat{i}+\hat{j}+\hat{k}$. It is:

Let the required unit vector be $\vec{n}$.Since $\vec{n}$ is coplanar to $\vec{a}$ and $\vec{b}$, it can be written as:$\vec{n}=x\vec{a}+y\vec{b}$$\vec{n}=x(\hat{i}+\hat{j}+2\hat{k})+y(\hat{i}+2\hat{j}+\hat{k})$$\vec{n}=(x+y)\hat{i}+(x+2y)\hat{j}+(2x+y)\hat{k}$ Since $\vec{n}\perp \vec{c}$:$\vec{n}\cdot \vec{c}=0⟹(x+y)(1)+(x+2y)(1)+(2x+y)(1)=0$$4x+4y=0⟹y=-x$ Substitute $y=-x$ in the expression for $\vec{n}$:$\vec{n}=0\hat{i}+(x-2x)\hat{j}+(2x-x)\hat{k}=-x\hat{j}+x\hat{k}=x(-\hat{j}+\hat{k})$ Since $\vec{n}$ is a unit vector, $|\vec{n}|=1$:$\sqrt{(-x{)}^2+x^2}=1⟹\sqrt{2x^2}=1⟹x\sqrt{2}=1⟹x=\pm \frac{1}{\sqrt{2}}$ So, $\vec{n}=\pm \frac{1}{\sqrt{2}}(-\hat{j}+\hat{k})$, which can be written as $\pm \left(\frac{\hat{j}-\hat{k}}{\sqrt{2}}\right)$.