Parametric equations of given circle is $x=12 \cos \theta, y=12 \sin \theta$
$\left[\because\right.$ Parametric equation of $x^2+y^2=r^2$
is $x=r \cos \theta, y=r \sin \theta$ ]
Now, coordinates of point $A$ are given by
$$
\begin{aligned}
& x=12 \cos \frac{\pi}{3}, y=12 \sin \frac{\pi}{3} \\
& \Rightarrow \quad x=12 \cdot \frac{1}{2}, y=12 \cdot \frac{\sqrt{3}}{2} \\
& \Rightarrow \quad x=6 ; y=6 \sqrt{3}
\end{aligned}
$$

$$
\text { i.e. } \quad A \equiv(6,6 \sqrt{3})
$$



and coordinates of point $B$ are given by
$$
\begin{array}{ll}
& x=12 \cos \frac{\pi}{6}, y=12 \sin \frac{\pi}{6} \\
\Rightarrow \quad & x=12 \cdot \frac{\sqrt{3}}{2} \cdot y=12 \cdot \frac{1}{2} \\
\Rightarrow \quad & x=6 \sqrt{3}, y=6 \\
\text { i.e. } \quad & B=(6 \sqrt{3}, 6)
\end{array}
$$
i.e. $\quad B \equiv(6 \sqrt{3}, 6)$
Clearly, length of chord
$$
\begin{aligned}
A B & =\sqrt{(6 \sqrt{3}-6)^2+(6-6 \sqrt{3})^2} \\
& =\sqrt{2 \times 6^2(\sqrt{3}-1)^2} \\
& =6 \sqrt{2}(\sqrt{3}-1) \\
& =6(\sqrt{6}-\sqrt{2})
\end{aligned}
$$