First Solution. Assume, for the sake of contradiction, that at least two of the numbers
$$\frac{2}{a} + \frac{3}{b} + \frac{6}{c}, \quad \frac{2}{b} + \frac{3}{c} + \frac{6}{a}, \quad \frac{2}{c} + \frac{3}{a} + \frac{6}{b}$$
are less than $6$. Without loss of generality, we may further assume that the first and the last are less than $6$. Then
$$\frac{5}{a} + \frac{9}{b} + \frac{8}{c} < 12.$$
Also, because $b + c \ge a(bc - 1)$, we have
$$\frac{1}{a} \ge \frac{bc - 1}{b + c}.$$
It follows that
$$\frac{5(bc - 1)}{b + c} + \frac{9}{b} + \frac{8}{c} < 12,$$
or
$$5b^{2}c^{2} + 12bc - 12b^{2}c - 12c^{2}b + 9c^{2} + 8b^{2} < 0. \quad (1)$$
Completing squares yields
$$(2bc - 2b - 3c)^2 + b^2(c - 2)^2 < 0,$$
a contradiction. This proves the conclusion. To obtain equalities, that is two of the members
$$\frac{2}{a} + \frac{3}{b} + \frac{6}{c}, \quad \frac{2}{b} + \frac{3}{c} + \frac{6}{a}, \quad \frac{2}{c} + \frac{3}{a} + \frac{6}{b}$$
are equal to $6$, we must have $c - 2 = 0$ and $2bc - 2b - 3c = 0$, that is $c = 2$, $b = 3$, $a = 1$. Therefore the equalities hold if and only if $(a, b, c)$ is one of the triples $(1, 3, 2)$, $(3, 2, 1)$, $(2, 1, 3)$.

Or: Rewriting (1) as a quadratic form in $b$ yields
$$(5c^2 - 12c + 8)b^2 - 12c(c - 1)b + 9c^2 < 0,$$
which is impossible, because the leading coefficient
$$5c^2 - 12c + 8 = 5\left(c - \frac{6}{5}\right)^2 + \frac{4}{5}$$
is always positive and the discriminant
$$\begin{aligned} \Delta &= [12c(c-1)]^2 - 36c^2(5c^2 - 12c + 8) \\ &= 36c^2[4(c-1)^2 - (5c^2 - 12c + 8)] = -36c^2(c-2)^2 \end{aligned}$$
is always nonpositive.

Second Solution. (by David Shin) Perform the substitutions $x = 1/a$, $y = 1/b$, and $z = 1/c$. It suffices to prove that at least two of the inequalities
$$2x + 3y + 6z \le 6, \quad 2y + 3z + 6x \le 6, \quad 2z + 3x + 6y \le 6$$
are true, where
$$x, y, z > 0 \quad \text{and} \quad xy + yz + zx \ge 1.$$
Assume, for the sake of contradiction, that at least two of the given inequalities are false. Without loss of generality, we may assume that $2x + 3y + 6z < 6$ and $2y + 3z + 6x < 6$. Then
$$\begin{aligned} & 144 > [(2x + 3y + 6z) + (2y + 3z + 6x)]^2 \\ &= (8x + 5y + 9z)^2 \\ &= 64x^2 + 80xy + 25y^2 + 81z^2 + 90yz + 144zx \\ &= 64x^2 - 64xy + 16y^2 + 9y^2 - 54yz + 81z^2 + 144(xy + yz + zx) \\ &= (8x - 4y)^2 + (3y - 9z)^2 + 144 \ge 144, \end{aligned}$$
a contradiction. Thus, our assumptions is false and at least two of the desired inequalities must be true. To obtain equalities, that is two of the numbers
$$\frac{2}{a} + \frac{3}{b} + \frac{6}{c}, \quad \frac{2}{b} + \frac{3}{c} + \frac{6}{a}, \quad \frac{2}{c} + \frac{3}{a} + \frac{6}{b}$$
are equal to $6$, we must have $8x - 4y = 3y - 9z = 0$, or $a : b : c = (1/x) : (1/y) : (1/z) = 2 : 1 : 3$. Therefore the equalities hold if and only if $(a, b, c) = (1, 3, 2)$, $(3, 2, 1)$, $(2, 1, 3)$.