Let the vector be given as $a \mathbf{i}+b \mathbf{j}+c \mathbf{k}$. For this vector to be coplanar with $\mathbf{i}+\mathbf{j}+2 \mathbf{k}$ and $\mathbf{i}+2 \mathbf{j}+\mathbf{k}$, we will have $a \mathbf{i}+b \mathbf{j}+c \mathbf{k}=p(\mathbf{i}+\mathbf{j}+2 \mathbf{k})+r(\mathbf{i}+2 \mathbf{j}+\mathbf{k})$
This gives, $a=p+r$
$\begin{aligned}
& b=p+2 r \\
& c=2 p+r
\end{aligned}$
For the vector $a \mathbf{i}+b \mathbf{j}+c \mathbf{k}$ to be perpendicular to $\mathbf{i}+\mathbf{j}+\mathbf{k}$, we will have $(a \mathbf{i}+b \mathbf{j}+c \mathbf{k}) .(\mathbf{i}+\mathbf{j}+\mathbf{k})=0$
$\Rightarrow a+b+c=0$
Adding equation (i) to (iii), we get $4 p+4 r=a+b+c$
$\Rightarrow 4(p+r)=0 \Rightarrow p=-r$
Now with the help of (i), (ii) and (iii), we get
$a=0, \quad b=r, \quad c=p=-r$
Hence the required vector is $r(\mathbf{j}-\mathbf{k})$
To be its unit vector
$r^2+r^2=1 \Rightarrow r= \pm \frac{1}{\sqrt{2}}$
Hence the required unit vector is,
$\pm \frac{1}{\sqrt{2}}(\mathbf{j}-\mathbf{k})$
Trick : Check for option(a) $\frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}}$ is a unit vector and perpendicular to $\mathbf{i}+\mathbf{j}+\mathbf{k}$ $\left|\begin{array}{ccc}\frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & 0 \\ 1 & 1 & 2 \\ 1 & 2 & 1\end{array}\right|=-\frac{4}{\sqrt{2}} \neq 0$
So it is not coplanar with the given vector.
(b), $\pm\left(\frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}}\right)$
Check for option
(b), $\pm\left(\frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}}\right)$ is a unit vector and also perpendicular to $\mathbf{i}+\mathbf{j}+\mathbf{k},\left|\begin{array}{ccc}0 & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ 1 & 1 & 2 \\ 1 & 2 & 1\end{array}\right|=0$
So, it is also coplanar with the given vectors.