If 3 and 4 are intercepts of a line $L \equiv 0$, then the distance of $L \equiv 0$ from the origin is

The intercept form of a line with x-intercept $a$ and y-intercept $b$ is given by $\dfrac{x}{a} + \dfrac{y}{b} = 1$.

Given $a = 3$ and $b = 4$, the equation of the line $L$ is $\dfrac{x}{3} + \dfrac{y}{4} = 1$.

Multiplying by 12, we get $4x + 3y = 12$, or $4x + 3y - 12 = 0$.

The distance $d$ of a line $Ax + By + C = 0$ from the origin $(0, 0)$ is given by $d = \dfrac{|C|}{\sqrt{A^2 + B^2}}$.

Substituting $A = 4$, $B = 3$, and $C = -12$, we get $d = \dfrac{|-12|}{\sqrt{4^2 + 3^2}} = \dfrac{12}{\sqrt{16 + 9}} = \dfrac{12}{\sqrt{25}} = \dfrac{12}{5}$.

Answer: $\dfrac{12}{5}$ units