Let the equation of line passing through origin is $y=m x$ which cut the parallel lines $x-y+10=0$ and $x-y+20=0$ at points $A$ and $B$ respectively. Then, co-ordinates of points $A$ and $B$ are,
$$
\begin{gathered}
y=m x \text { and } x-y+10=0 \\
x-m x=-10 \\
x=\frac{-10}{1-m}=\frac{10}{m-1} \text { and } y=\frac{10 m}{m-1} \\
\therefore A\left(\frac{10}{m-1}, \frac{10 m}{m-1}\right)
\end{gathered}
$$
So,
$$
\begin{aligned}
O A & =\sqrt{\frac{100}{(m-1)^2}+\frac{100 m^2}{(m-1)^2}} \\
& =\sqrt{\frac{100\left(1+m^2\right)}{(m-1)^2}}=\frac{10 \sqrt{\left(1+m^2\right)}}{m-1}
\end{aligned}
$$

Similarly, coordinate of point $B$ is
$\therefore$
$$
B\left(\frac{20}{m-1}, \frac{20 m}{m-1}\right)
$$

Then, $O B=\sqrt{\frac{400}{(m-1)^2}+\frac{400 m^2}{(m-1)^2}}=\frac{20 \sqrt{1+m^2}}{(m-1)}$
Let point $P(h, k)$ lies on the line $y=m x$
So,
$$
k=m h \Rightarrow m=\frac{k}{h}
$$
and $\quad O P=\sqrt{h^2+k^2}$
Now, $O A, O P, O B$ in H.P.
So, $\frac{1}{O A}, \frac{1}{O P}, \frac{1}{O B}$ in AP.
Hence, $\quad \frac{1}{O P}=\frac{1}{O A}+\frac{1}{O B}$
$$
\begin{aligned}
& \frac{2}{\sqrt{h^2+k^2}}=\frac{m-1}{10 \sqrt{1+m^2}}+\frac{m-1}{20 \sqrt{1+m^2}} \\
& \frac{2}{\sqrt{h^2+k^2}}=\frac{m-1}{10 \sqrt{1+m^2}}\left[1+\frac{1}{2}\right]
\end{aligned}
$$

Putting the value of $m$, we get
$$
\begin{aligned}
& \frac{2}{\sqrt{h^2+k^2}}=\frac{\left(\frac{k}{h}-1\right)}{10 \cdot \sqrt{1+\frac{k^2}{h^2}}} \cdot \frac{3}{2} \\
\Rightarrow \quad \frac{2}{\sqrt{h^2+k^2}} & =\frac{3(k-h) / h}{20 \frac{\sqrt{h^2+k^2}}{h}} \\
\Rightarrow \quad \frac{2}{\sqrt{h^2+k^2}}= & \frac{3(k-h)}{20 \sqrt{h^2+k^2}} \\
\Rightarrow \quad 3(k-h) & =40 \Rightarrow 3 h-3 k+40=0
\end{aligned}
$$

Hence, locus of the point is
$$
3 x-3 y+40=0 .
$$