Find Io, I1, I2 when (a) Switch is closed for balanced load (b) Switch is open.

The three $5A$ currents are the line currents ($I_L$) entering a balanced three-phase delta-connected load.  
The currents $I_0, I_1, I_2$ are the phase currents ($I_P$) within the delta load.

For a balanced delta-connected load, the magnitude of the line current is $\sqrt{3}$ times the magnitude of the phase current:
$$
I_L = \sqrt{3} I_P
$$
Therefore, the phase current is:
$$
I_P = \frac{I_L}{\sqrt{3}}
$$

Given $I_L = 5A$:
$$
I_0 = I_1 = I_2 = \frac{5}{\sqrt{3}} \text{ A}
$$
$$
I_0 = I_1 = I_2 \approx \frac{5}{1.732} \text{ A}
$$
$$
I_0 = I_1 = I_2 \approx 2.887 \text{ A}
$$
The currents are:
$$
\boxed{I_0 = I_1 = I_2 = \frac{5}{\sqrt{3}} \text{ A}}
$$

When switch S is open, the third input line (the bottom one) is disconnected from the load. This means the current in that line becomes $0A$. The system is no longer a balanced three-phase system; it becomes an open-delta (or V-connection) system.

In an open-delta system, the two remaining line currents are equal to the phase currents in the two active phases. The third phase current (corresponding to the disconnected line) will be zero.  
From the diagram, the top input line (corresponding to $I_0$) and the middle input line (corresponding to $I_1$) are still connected, each carrying $5A$. The bottom input line (corresponding to $I_2$) is disconnected.  
Therefore, the phase currents will be:
$$
I_0 = 5 \text{ A}
$$
$$
I_1 = 5 \text{ A}
$$
$$
I_2 = 0 \text{ A}
$$
The currents are:
$$
\boxed{I_0 = 5 \text{ A}, I_1 = 5 \text{ A}, I_2 = 0 \text{ A}}
$$