A unit vector in the plane of $(\mathbf{i}+2\mathbf{j}+\mathbf{k})$ and $(\mathbf{i}+\mathbf{j}+2\mathbf{k})$ and perpendicular to $(2\mathbf{i}+\mathbf{j}+\mathbf{k})$.

Let, $\mathbf{a}=(\mathbf{i}+2\mathbf{j}+\mathbf{k})$

and $\begin{array}{r}\mathbf{b}=(\mathbf{i}+\mathbf{j}+2\mathbf{k})\\ \mathbf{c}=(2\mathbf{i}+\mathbf{j}+\mathbf{k})\end{array}$

and r be a vector

such that $\mathbf{r}=(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})$.

It is given that r, a and b lie in the same plane, so

$\begin{array}{l}\mathbf{r}\cdot (\mathbf{a}\times \mathbf{b})=0\\ \Rightarrow \left|\begin{array}{lll}x& y& z\\ 1& 2& 1\\ 1& 1& 2\end{array}\right|=0\end{array}$

$\Rightarrow 3x-y-z=0$

Also,

$\begin{array}{l}\mathbf{r}\cdot \mathbf{c}=0\\ \Rightarrow 2x+y+z=0\end{array}$

Solving Eqs. (i) and (ii), we get 

$\begin{array}{l}\frac{x}{0}=\frac{y}{-5}=\frac{z}{5}\\ \frac{x}{0}=\frac{y}{-1}=\frac{z}{1}=\lambda \left(\text{ say }\right)\end{array}$

Therefore, $\widehat{r}=\frac{\mathbf{r}}{\left|\mathbf{r}\right|}$

$\Rightarrow \widehat{r}=\pm \left(\frac{\mathbf{j}\mathbf{-}\mathbf{k}}{\sqrt{\mathbf{2}}}\right)$