Let a variable line $L$ meet $x$-axis and $y$-axis at points A and B, respectively. Suppose the distance of the line $L$ from the origin is $3$ units. Then the equation of the locus of the point C that divides the line segment AB internally in the ratio $2 : 1$ is

Let the equation of the variable line $L$ be $\dfrac{x}{a} + \dfrac{y}{b} = 1$.

The line meets the $x$-axis at $A(a, 0)$ and the $y$-axis at $B(0, b)$.

The perpendicular distance from the origin $(0,0)$ to the line $L$ is given as $3$ units. Using the perpendicular distance formula:

$\dfrac{|-1|}{\sqrt{\dfrac{1}{a^2} + \dfrac{1}{b^2}}} = 3$

Squaring both sides, we get:

$\dfrac{1}{a^2} + \dfrac{1}{b^2} = \dfrac{1}{9}$

Let $C(h, k)$ be the point that divides the line segment $AB$ internally in the ratio $2 : 1$. This means $AC : CB = 2 : 1$.

Using the section formula, the coordinates of $C$ are:

$h = \dfrac{2(0) + 1(a)}{2 + 1} = \dfrac{a}{3} \Rightarrow a = 3h$

$k = \dfrac{2(b) + 1(0)}{2 + 1} = \dfrac{2b}{3} \Rightarrow b = \dfrac{3k}{2}$

Substituting the values of $a$ and $b$ into the distance condition:

$\dfrac{1}{(3h)^2} + \dfrac{1}{\left(\dfrac{3k}{2}\right)^2} = \dfrac{1}{9}$

$\dfrac{1}{9h^2} + \dfrac{4}{9k^2} = \dfrac{1}{9}$

Multiplying the entire equation by $9$, we get:

$\dfrac{1}{h^2} + \dfrac{4}{k^2} = 1$

Replacing $(h, k)$ with $(x, y)$, the equation of the locus of $C$ is:

$\dfrac{1}{x^2} + \dfrac{4}{y^2} = 1$

Answer: $\dfrac{1}{x^2} + \dfrac{4}{y^2} = 1$