Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
From the given data, we can see that $(\varepsilon_0)_2$ leads the $(\varepsilon_0)_1$ by an angle of 120$^\circ$, while $(\varepsilon_0)_3$ lags the $(\varepsilon_0)_1$ by an angle of 120$^\circ$ as well.
See the three phasors in the figure above.
$$\color{blue}{\bf [b]}$$
The three vectors, in the first figure below, have the same length and if we connect them head to tail, they will form a triangle which means that $(\varepsilon_0)_3$ head is at the tail of $(\varepsilon_0)_1$.
Hence, the resultant here is zero, which means that the sum of the three phases is zero.
See the second figure below.
$$\color{blue}{\bf [c]}$$
To show that the potential difference between any two of the phases has the rms value, we need to use the third figure below.
We need to find the value of $(\varepsilon_0)'$.
From the geometry of the third figure below,
$$\dfrac{(\varepsilon_0)'}{\sin120^{\circ}}=\dfrac{ \varepsilon_0 }{\sin30^{\circ}}$$
Hence,
$$ (\varepsilon_0)' =\dfrac{ \sin(120^{\circ})\varepsilon_0 }{\sin(30^{\circ})}=\boxed{\sqrt{3}\;\varepsilon_0 }$$
