Step 1: Set up the intercept points.
Let the line meet the $x$-axis at $A = (a, 0)$ and the $y$-axis at $B = (0, b)$. Its equation in intercept form is $\frac{X}{a} + \frac{Y}{b} = 1$.
Step 2: Use the distance-from-origin condition.
The perpendicular distance from the origin to this line is $3$, so $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = 3$, which gives $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{9}$.
Step 3: Locate point C dividing AB in ratio $2 : 1$.
Using the section formula with C dividing from A to B as $2 : 1$, $x = \frac{2(0) + 1(a)}{3} = \frac{a}{3}$ and $y = \frac{2(b) + 1(0)}{3} = \frac{2b}{3}$.
Step 4: Express $a$ and $b$ through $x$ and $y$.
From the above, $a = 3x$ and $b = \frac{3y}{2}$.
Step 5: Substitute into the distance condition.
Replace in $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{9}$: $\frac{1}{9x^2} + \frac{1}{\frac{9y^2}{4}} = \frac{1}{9}$, that is $\frac{1}{9x^2} + \frac{4}{9y^2} = \frac{1}{9}$.
Step 6: Clear the common factor.
Multiply every term by $9$: $\frac{1}{x^2} + \frac{4}{y^2} = 1$. This is the locus of C.
\[ \boxed{\dfrac{1}{x^2} + \dfrac{4}{y^2} = 1} \]