Length of Chord Intersecting Circle in First Quadrant

Let the centre of the circle $C$ be $(h, k)$ and its radius be $R$.

Since the circle lies in the first quadrant and touches the $x$-axis at a distance of $3$ units from the origin, its point of contact with the $x$-axis is $(3, 0)$.

Therefore, the $x$-coordinate of the centre is $h = 3$, and the radius of the circle is equal to the $y$-coordinate of the centre, i.e., $R = k$ (where $k > 0$).

Thus, the centre of the circle is $(3, k)$ and its equation is: $(x - 3)^2 + (y - k)^2 = k^2$

The length of the intercept made by the circle on the $y$-axis is given by $2\sqrt{R^2 - h^2}$. We are given that this intercept length is $6\sqrt{3}$: $2\sqrt{k^2 - 3^2} = 6\sqrt{3}$

Dividing both sides by $2$: $\sqrt{k^2 - 9} = 3\sqrt{3}$

Squaring both sides: $k^2 - 9 = 27 \implies k^2 = 36$

Since the centre is in the first quadrant, $k > 0$, so $k = 6$.

Thus, the centre of the circle $C$ is $(3, 6)$ and its radius is $R = 6$.

Now, we need to find the length of the chord of the circle on the line $x - y = 3$, which can be rewritten as: $x - y - 3 = 0$

The perpendicular distance $d$ from the centre $(3, 6)$ to the line $x - y - 3 = 0$ is: $d = \frac{|3 - 6 - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|-6|}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$

The length of the chord $L$ intercepted by the circle on the line is given by: $L = 2\sqrt{R^2 - d^2}$

Substituting $R = 6$ and $d = 3\sqrt{2}$: $L = 2\sqrt{6^2 - (3\sqrt{2})^2} = 2\sqrt{36 - 18} = 2\sqrt{18} = 2 \times 3\sqrt{2} = 6\sqrt{2}$

Hence, the length of the chord is $6\sqrt{2}$.

Correct Option: C