A point charge q _ 1 = 4 q _ 0 is placed at origin. Another point...

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

$\dfrac{1}{4\pi{ε}_{0}}\dfrac{4{q}_{0}}{{x}^{2}} = \dfrac{1}{4\pi{ε}_{0}}\dfrac{{q}_{0}}{{\left(x - 12\right)}^{2}} ⇒ 2\left(x - 12\right) = x ⇒ x = 24cm$

Distance of proton from origin $= 24cm$.