Determine the accelerations a_1 , a_2 , and a_3 of the three blocks...

First, we find the maximum possible friction forces at each interface.

Between the $2 \text{ kg}$ and $3 \text{ kg}$ blocks:
f_{1,\text\max} = \mu_1 m_1 g = 0.2 \times 2 \times 10 = 4 \text{ N}

Between the $3 \text{ kg}$ and $7 \text{ kg}$ blocks:
f_{2,\text\max} = \mu_2 (m_1 + m_2) g = 0.3 \times (2 + 3) \times 10 = 15 \text{ N}

Between the $7 \text{ kg}$ block and the ground:
f_{3,\text\max} = \mu_3 (m_1 + m_2 + m_3) g = 0 \times 12 \times 10 = 0 \text{ N}

Let us assume all three blocks move together with a common acceleration $a$.
$a = \dfrac{F}{m_1 + m_2 + m_3} = \dfrac{10}{2 + 3 + 7} = \dfrac{10}{12} = \dfrac{5}{6} \text{ m/s}^2$

The required friction force $f_1$ to keep the $2 \text{ kg}$ block moving with this acceleration is:
$F - f_1 = m_1 a \Rightarrow 10 - f_1 = 2 \times \dfrac{5}{6} \Rightarrow f_1 = 10 - \dfrac{5}{3} = \dfrac{25}{3} \approx 8.33 \text{ N}$

Since f_1 > f_{1,\text\max} ($8.33 \text{ N} > 4 \text{ N}$), the $2 \text{ kg}$ block will slip relative to the $3 \text{ kg}$ block. Thus, kinetic friction $f_1 = 4 \text{ N}$ acts between them.

The acceleration of the $2 \text{ kg}$ block is:
$a_1 = \dfrac{F - f_1}{m_1} = \dfrac{10 - 4}{2} = 3 \text{ m/s}^2$

Now, the friction force $f_1 = 4 \text{ N}$ acts in the forward direction on the $3 \text{ kg}$ block. We check if the $3 \text{ kg}$ and $7 \text{ kg}$ blocks move together.
The maximum friction available between them is f_{2,\text\max} = 15 \text{ N}. Since the driving force on the $3 \text{ kg}$ block is only $4 \text{ N}$, which is less than $15 \text{ N}$, they will not slip relative to each other and will move together.

The common acceleration of the $3 \text{ kg}$ and $7 \text{ kg}$ blocks is:
$a_2 = a_3 = \dfrac{f_1}{m_2 + m_3} = \dfrac{4}{3 + 7} = \dfrac{4}{10} = 0.4 \text{ m/s}^2$

(To verify, the friction force $f_2$ required between them is $m_3 a_3 = 7 \times 0.4 = 2.8 \text{ N}$, which is well below the maximum limit of $15 \text{ N}$).

Therefore, $a_1 = 3 \text{ m/s}^2$, $a_2 = 0.4 \text{ m/s}^2$, and $a_3 = 0.4 \text{ m/s}^2$.

Answer: $a_1 = 3 \text{ m/s}^2$, $a_2 = 0.4 \text{ m/s}^2$, $a_3 = 0.4 \text{ m/s}^2$