Given points $A(k, 2 k), B(3 k, 3 k)$ and $C(3,1)$ are collinear.
$$
\begin{array}{rlrl}
& \therefore & \text { Slope of } A B & =\text { Slope of } B C \\
& \therefore \quad \frac{2 k-2 k}{3 k-k} & =\frac{1-3 k}{3-3 k} \\
\Rightarrow & \frac{k}{2 k} & =\frac{1-3 k}{3-3 k} \Rightarrow 3-3 k=2(1-3 k) \\
\Rightarrow & 1 & =-3 k \Rightarrow k=-\frac{1}{3}
\end{array}
$$
$\therefore$ Given points become $A\left(-\frac{1}{3},-\frac{2}{3}\right), B(-1,-1)$ and $C(3,1)$.
$\therefore$ Equation of line passing through $B$ and $C$ is
$$
\begin{array}{rlrl}
y+1 & =\frac{1+1}{3+1}(x+1) \\
\Rightarrow & & y+1 & =\frac{2}{4}(x+1) \\
\Rightarrow & 2(y+1) & =(x+1) \\
\Rightarrow & 2 y-x+1 & =0
\end{array}
$$
Now, the distance from $(0,0)$ to the above line is
$$
d=\frac{|2(0)-0+1|}{\sqrt{2^2+1^2}} \Rightarrow=\frac{1}{\sqrt{4+1}}=\frac{1}{\sqrt{5}}
$$
Alternate Solution
Equation of line passing through $(k, 2 k)$ and $(3 k, 3 k)$ is
$$
\begin{array}{rlrl}
& (y-2 k) & =\frac{3 k-2 k}{3 k-k}(x-k) \\
\Rightarrow & y-2 k & =\frac{k}{2 k}(x-k) \\
\Rightarrow \quad y-2 k & =\frac{1}{2}(x-k)
\end{array}
$$
Since, above line is passing through $(1,1)$.
$$
\begin{array}{rlrl}
& \therefore & 1-2 k & =\frac{1}{2}(1-k) \\
\Rightarrow & 2-4 k & =1-k \\
\Rightarrow & 1 & =3 k \Rightarrow k=\frac{1}{3}
\end{array}
$$
On putting $k=\frac{1}{3}$ in Eq. (i), we get
$$
\begin{aligned}
y-\frac{2}{3} & =\frac{1}{2}\left(x-\frac{1}{3}\right) \\
\Rightarrow \quad 3 y-2 & =\frac{1}{2}(3 x-1) \\
\Rightarrow \quad 6 y-4 & =3 x-1 \\
\Rightarrow \quad 6 y-3 x-3 & =0
\end{aligned}
$$
or $2 y-x-1=0$
$\therefore$ Perpendicular distance from $(0,0)$ to the above line is
$$
\begin{aligned}
d & =\frac{|2(0)-0-1|}{\sqrt{2^2+1^2}}=\frac{1}{\sqrt{4+1}} \\
& =\frac{1}{\sqrt{5}}
\end{aligned}
$$