By the given condition, we have
$$a^2 + (2b - 1)^2 + (3c - 2)^2 = 3.$$
By making use of the Cauchy inequality, we get
$$3[a^2 + (2b - 1)^2 + (3c - 2)^2] \geq (a + 2b - 1 + 3c - 2)^2,$$
namely, $(a + 2b + 3c - 3)^2 \leq 9$. Therefore,
$$a + 2b + 3c \leq 6.$$
Again, by the Cauchy inequality we get
$$\left( \frac{1}{a} + \frac{2}{b} + \frac{3}{c} \right) (a + 2b + 3c) \geq (1 + 2 + 3)^2.$$
Thus,
$$\frac{1}{a} + \frac{2}{b} + \frac{3}{c} \ge \frac{36}{a + 2b + 3c} \ge 6,$$
and the equal sign holds when $a = b = c = 1$.
Therefore, the minimum value of $\frac{1}{a} + \frac{2}{b} + \frac{3}{c}$ is $6$.