Let the unit vector be
$\begin{aligned}
& \mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \\
& \text {and } \mathbf{a}=\mathbf{i}+\mathbf{j}+3 \mathbf{k}, \mathbf{b}=\mathbf{i}+3 \mathbf{j}+\mathbf{k} \\
& \text {and } \mathbf{c}=\mathbf{i}+\mathbf{j}+\mathbf{k}
\end{aligned}$
Given, $[\mathbf{r}, \mathbf{a} \mathbf{b}]=0$, i.e., coplanar.
$\begin{array}{ccc}
\Rightarrow & \left|\begin{array}{ccc}
x & y & z \\
1 & 1 & 3 \\
1 & 3 & 1
\end{array}\right|=0 \\
\Rightarrow & x(1-9)-y(1-3)+z(3-1)=0 \\
\Rightarrow & -8 x+2 y+2 x=0 \\
\Rightarrow & -4 x+y+z=0
\end{array}$
and $\quad \mathbf{r} \cdot \mathbf{c}=0$, i.e., perpendicular
$\begin{array}{cc}
\Rightarrow & (x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) \cdot(\mathbf{i}+\mathbf{j}+\mathbf{k})=0 \\
\Rightarrow & x+y+z=0
\end{array}$
On solving Eqs. (i) and (ii), we get
$\begin{aligned}
5 y+5 z & =0 \\
y & =-z
\end{aligned}$
$\because \mathbf{r}$ is a unit vector.
$\begin{array}{ll}
\therefore & |\mathbf{r}|=1=\sqrt{x^2+y^2+z^2} \\
\Rightarrow & x^2+y^2+z^2=1 \\
\Rightarrow & x^2+2 y^2=1 \text { [from Eq. (iii)]. }
\end{array}$
Put $y=-z$ in Eq. (i), we get
$-4 x=0 \Rightarrow x=0$
From Eq. (iv), we get
$2 y^2=1 \Rightarrow y= \pm \frac{1}{\sqrt{2}}$
Required vector is
$\begin{aligned}
\mathbf{r} & =x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \\
& =0 \mathbf{i} \mp \frac{1}{\sqrt{2}} \mathbf{j} \pm \frac{1}{\sqrt{2}} \mathbf{k} \\
& =\frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}} \text { or } \frac{-\mathbf{j}+\mathbf{k}}{\sqrt{2}}
\end{aligned}$