Let a proton having charge $q_0$ on the $x$ axis at distance $x$ from $q_2$ and $(12+x)$ distance from $q_1$.

Now, balance the force between them $\vec{F}_1+\vec{F}_2=0$

$$ \frac{K 4 q_0\left(q_0\right)}{(12+x)^2}+\frac{K\left(-q_0\right)\left(q_0\right)}{x^2}=0 $$

$$ \Rightarrow $$ $$ \frac{4 K q_0^2}{(12+x)^2}=\frac{K\left(q_0\right)^2}{x^2} $$

$$ \Rightarrow $$ $$ \frac{4}{(12+x)^2}=\frac{1}{x^2} $$

$$ \Rightarrow $$ $$ \frac{2}{12+x}=\frac{1}{x} $$

$$ \Rightarrow $$ $$ 2 x=12+x $$

$$ \Rightarrow $$ $$ x=12 \mathrm{~cm} $$

The charge $q_0$ is at distance of $24 \mathrm{~cm}$ from charge $q_1$ on $x$-axis.

$$ \therefore $$ Distance from origin is $12+12=24 \mathrm{~cm}$