In the arrangement depicted in the figure, let m_1 = 1 kg , m_2 = 2...

Let $T_1$ be the tension in the string connected to $m_1$ and $T_2$ be the tension in the string connecting $m_2$ and $m_3$. Since the lower pulley is massless, $T_1 = 2T_2$.

Let $a_1$ be the upward acceleration of $m_1$. The lower pulley will have a downward acceleration of $a_1$.
Let $a'$ be the downward acceleration of $m_3$ relative to the lower pulley. Then $m_2$ has an upward acceleration $a'$ relative to the lower pulley.
The absolute downward accelerations of $m_2$ and $m_3$ are $a_2 = a_1 - a'$ and $a_3 = a_1 + a'$.

Writing the equations of motion for each mass:
For $m_1$: $T_1 - m_1 g = m_1 a_1$
$\Rightarrow 2T_2 - g = a_1 \Rightarrow T_2 = \dfrac{a_1 + g}{2}$

For $m_2$: $m_2 g - T_2 = m_2 a_2$
$\Rightarrow 2g - T_2 = 2(a_1 - a') \Rightarrow g - \dfrac{T_2}{2} = a_1 - a'$

For $m_3$: $m_3 g - T_2 = m_3 a_3$
$\Rightarrow 3g - T_2 = 3(a_1 + a') \Rightarrow g - \dfrac{T_2}{3} = a_1 + a'$

Adding the equations for $m_2$ and $m_3$ to eliminate $a'$:
$2g - T_2\left(\dfrac{1}{2} + \dfrac{1}{3}\right) = 2a_1$
$2g - \dfrac{5}{6}T_2 = 2a_1$

Substituting $T_2 = \dfrac{a_1 + g}{2}$ into the above equation:
$2g - \dfrac{5}{6}\left(\dfrac{a_1 + g}{2}\right) = 2a_1$
$2g - \dfrac{5}{12}g = 2a_1 + \dfrac{5}{12}a_1$
$\dfrac{19}{12}g = \dfrac{29}{12}a_1 \Rightarrow a_1 = \dfrac{19g}{29}$ (upwards)

Now, finding $T_2$:
$T_2 = \dfrac{\dfrac{19g}{29} + g}{2} = \dfrac{24g}{29}$

Finding the accelerations $a_2$ and $a_3$:
$a_2 = g - \dfrac{T_2}{2} = g - \dfrac{12g}{29} = \dfrac{17g}{29}$ (downwards)
$a_3 = g - \dfrac{T_2}{3} = g - \dfrac{8g}{29} = \dfrac{21g}{29}$ (downwards)

To find the time elapsed before $m_1$ strikes the upper pulley, we use the kinematic equation $s = ut + \dfrac{1}{2}at^2$.
Given $s = 20 \text{ cm} = 0.2 \text{ m}$, $u = 0$, and $a_1 = \dfrac{19g}{29}$.
$0.2 = \dfrac{1}{2} \times \left(\dfrac{19 \times 9.8}{29}\right) \times t^2$
$0.4 = \dfrac{186.2}{29} t^2$
$t^2 = \dfrac{0.4 \times 29}{186.2} \approx 0.0623$
$t \approx 0.25 \text{ s}$

Answer: $\dfrac{19g}{29}$ (up), $\dfrac{17g}{29}$ (down), $\dfrac{21g}{29}$ (down), $0.25 \text{ s}$