A right circular cone has a vertical height of $(3 - \sqrt{3})$ cm and a slant height of $l$ cm. The volume of the cone is $(10 + 2\sqrt{3})\pi$ cm$^3$. Without using a calculator, express $l^2$ in the form $a + b\sqrt{3}$, where $a$ and $b$ are constants. [Volume of cone = $\frac{1}{3}\pi r^2 h$]

1. Set up the equation for the volume of the cone to find $r^2$.

Volume $= \frac{1}{3}\pi r^2 h$
$(10 + 2\sqrt{3})\pi = \frac{1}{3}\pi r^2 (3 - \sqrt{3})$
$10 + 2\sqrt{3} = \frac{1}{3} r^2 (3 - \sqrt{3})$
$30 + 6\sqrt{3} = r^2 (3 - \sqrt{3})$

2. Express $r^2$ in terms of surds and rationalise the denominator.

$r^2 = \frac{30 + 6\sqrt{3}}{3 - \sqrt{3}}$
$r^2 = \frac{(30 + 6\sqrt{3})(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})}$

3. Expand and simplify the expression for $r^2$.

$r^2 = \frac{90 + 30\sqrt{3} + 18\sqrt{3} + 18}{3^2 - (\sqrt{3})^2}$
$r^2 = \frac{108 + 48\sqrt{3}}{9 - 3}$
$r^2 = \frac{108 + 48\sqrt{3}}{6}$
$r^2 = 18 + 8\sqrt{3}$

4. Use Pythagoras' theorem to relate $l^2$, $r^2$, and $h^2$.

$l^2 = r^2 + h^2$
$l^2 = (18 + 8\sqrt{3}) + (3 - \sqrt{3})^2$

5. Expand $h^2$ and simplify to find $l^2$.

$(3 - \sqrt{3})^2 = 9 - 6\sqrt{3} + 3 = 12 - 6\sqrt{3}$
$l^2 = 18 + 8\sqrt{3} + 12 - 6\sqrt{3}$
$l^2 = 30 + 2\sqrt{3}$