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In $\Delta ABC$, the angles are:
$\angle A = 105^\circ$
$\angle B = 15^\circ$
$\angle C = 180^\circ - (105^\circ + 15^\circ) = 60^\circ$

Given, $c = \sqrt{3} - 1$.
Using the Sine Rule:
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Here,
$\sin 60^\circ = \frac{\sqrt{3}}{2}$
$\sin 105^\circ = \sin(60^\circ + 45^\circ) = \frac{\sqrt{3}+1}{2\sqrt{2}}$
$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \frac{\sqrt{3}-1}{2\sqrt{2}}$

Now, solving for $a$:
$a = \frac{c \sin A}{\sin C} = \frac{(\sqrt{3}-1) \cdot \frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}}{2}} = \frac{3-1}{\sqrt{6}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}$

Solving for $b$:
$b = \frac{c \sin B}{\sin C} = \frac{(\sqrt{3}-1) \cdot \frac{\sqrt{3}-1}{2\sqrt{2}}}{\frac{\sqrt{3}}{2}} = \frac{(\sqrt{3}-1)^2}{\sqrt{6}} = \frac{4 - 2\sqrt{3}}{\sqrt{6}} = \frac{2\sqrt{6} - 3\sqrt{2}}{3}$

Therefore, the sides of the triangle are:
$a = \frac{\sqrt{6}}{3}, \quad b = \frac{2\sqrt{6}-3\sqrt{2}}{3}, \quad c = \sqrt{3}-1$