Answer. Equality occurs if and only if $a = b = c = 2$.

We set $x = a + 1$, $y = b + 1$ and $z = c + 1$. Thus we have to show
$$xyz \geq 27$$
subject to
$$xyz = xy + yz + zx.$$
From the constraint we get
$$xyz = xy + yz + zx \geq 3\sqrt[3]{x^2y^2z^2}$$
by using the inequality between the arithmetic and the geometric means of $xy$, $yz$ and $zx$. This is clearly equivalent to $xyz \geq 27$.
Equality occurs if and only if $xy = yz = zx$, or, equivalently, $x = y = z$. By the constraint, this is equivalent to $x = y = z = 3$ and finally $a = b = c = 2$.