The admittance of this circuit can be written as :
Y = \$\dfrac{1}{sL} + \dfrac{1}{R + \dfrac{1}{sC}}\ = \dfrac{CLs^2 + CRs + 1}{Ls(CRs + 1)}\$.
Substituting \$s = j\omega\$, multiplying the denominator by its complex cojugated and simplifying into real and imaginary parts gives us:
\$\dfrac{R}{\frac{1}{C^2 \omega^2} + R^2} + j\left(\dfrac{1}{C\omega\left(\frac{1}{C^2\omega^2} + R^2 \right)} - \dfrac{1}{L\omega}\right)\$.
A complex admittance consists of a conductance (real part) and a susceptance (imaginary part).
Substituting the value of the resistance and frequency, we want
\$\dfrac{50}{\frac{1}{C^2 (2*\pi*10^9)^2} + 50^2} = 10^{-3}S.\$
Solving for C gives C \$\approx\$ 0.73 pF.
Plugging this value of C and R into
\$j\left(\dfrac{1}{C\omega\left(\frac{1}{C^2\omega^2} + R^2 \right)} - \dfrac{1}{L\omega}\right)\ = -j10^{-3}\$
and solving for L gives L \$\approx\$ 30 nH.
The admittance of this inductance is \$\approx 5.3*10^{-9}\$ S.
Looking here as a reference, the equation for the length of an open-circuited transmission line to act as an inductor is:
l = \${\frac {1}{\beta }}\left[\pi(n+1) -\operatorname{arccot} \left({\frac {\omega L}{Z_{0}}}\right)\right]\$, where \$L = 30*10^{-9}\$\$, \beta = \dfrac{2\pi f }{c_l}\$, \$f = 10^9\$, and \$c_{l} \approx 0.8*3.0*10^9 \frac{m}{s}\$.
Best Answer
This discussion will be very basic and pictorial. It will only discuss the star or wye connection and ignore delta (for now anyway).
Figure 1. This animation of a simple 3-phase power system shows the basic principle of a balanced load. Note how the particles' of current entering and leaving the star node sum to zero. Source: BillC at Wikimedia.
It should be clear that we could connect the star point of the generator to the star point of the load with a neutral conductor but that no current would flow in it as the three phases are perfectly balanced.
Figure 2. Taking the OP's diagram to represent the currents in the balanced load we can see that at examples (1), (2) and (3) that they sum to zero. (I didn't cheat by stretching and of the arrows in each set.)