A point charge q 1 = 4 q 0 is placed at origin. Another point charge q 2 = - q 0 is placed at x = 12  …

Let at point $P$ net electric field is zero.

As direction of electric field due to both charges at point $P$ is opposite, equating magnitude we get

$$\begin{gathered} \frac{1}{4\pi {\epsilon }_0}\frac{4q_0}{x^2}=\frac{1}{4\pi {\epsilon }_0}\frac{q_0}{{\left(x-12\right)}^2} \\ \Rightarrow 2\left(x-12\right)=x \\ \Rightarrow x=24 \mathrm{cm} \end{gathered}$$

Distance of proton from origin $=24 \mathrm{cm}$.