Let C be a circle having centre in the first quadrant and touching...

Let the centre of the circle be $(3, r)$ since it touches the $x$-axis at $(3, 0)$ and lies in the first quadrant.

The radius of the circle is $r$.

The length of the intercept on the $y$-axis is given by $2\sqrt{r^2 - d^2}$, where $d$ is the distance of the centre from the $y$-axis.

Here, $d = 3$.

Given that the intercept on the $y$-axis is $6\sqrt{3}$:

$2\sqrt{r^2 - 3^2} = 6\sqrt{3}$

$\sqrt{r^2 - 9} = 3\sqrt{3}$

$r^2 - 9 = 27$

$r^2 = 36 \Rightarrow r = 6$

The centre of the circle is $(3, 6)$ and its radius is $6$.

The given line is $x - y - 3 = 0$.

The perpendicular distance $p$ from the centre $(3, 6)$ to the line is:

$p = \dfrac{|3 - 6 - 3|}{\sqrt{1^2 + (-1)^2}} = \dfrac{6}{\sqrt{2}} = 3\sqrt{2}$

The length of the chord is $2\sqrt{r^2 - p^2}$.

Length of the chord $= 2\sqrt{36 - (3\sqrt{2})^2} = 2\sqrt{36 - 18} = 2\sqrt{18} = 6\sqrt{2}$

Answer: $6\sqrt{2}$