Alternate Circuit

Applying Kirchoff's laws to the circuit above gives three equations for the unknown currents.

$-V_A + Z_AI_A + \left(I_A+I_B+I_C\right)Z_N = 0$ $-V_B + Z_BI_B + \left(I_A+I_B+I_C\right)Z_N = 0$ $-V_C + Z_CI_C + \left(I_A+I_B+I_C\right)Z_N = 0$

In a matrix form, the equations can be rearranged as:

$\left[\begin{matrix}Z_N+Z_A&Z_N&Z_N\\Z_N&Z_N+Z_B&Z_N\\Z_N&Z_N&Z_N+Z_C\end{matrix}\right]\left[\begin{matrix}I_A\\I_B\\I_C\end{matrix}\right]=\left[\begin{matrix}V_A\\V_B\\V_C\end{matrix}\right]\implies ZI = V \implies I = Z^{-1}V$

Let:

$I_T = I_A + I_B + I_C=\left[\begin{matrix}1&1&1\end{matrix}\right]\left[\begin{matrix}I_A\\I_B\\I_C\end{matrix}\right]$

Having obtained the currents, the next step is to solve for the impedances $Z'_B$ , $Z'_C$ , $Z'_N$ . This can be done by using the same set of equations obtained using Kirchoff's laws but treating three impedances as unknown variables. Rearranging the equations in a matrix form again gives:

$\left[\begin{matrix}0&0&I_T\\I_B&0&I_T\\0&I_C&I_T\end{matrix}\right]\left[\begin{matrix}Z'_B\\Z'_C\\Z'_N\end{matrix}\right]=\left[\begin{matrix}V_A-I_AZ'_A\\V_B\\V_C\end{matrix}\right]$

$\implies \left[\begin{matrix}Z'_B\\Z'_C\\Z'_N\end{matrix}\right] = \left[\begin{matrix}0&0&I_T\\I_B&0&I_T\\0&I_C&I_T\end{matrix}\right]^{-1}\left[\begin{matrix}V_A-I_AZ'_A\\V_B\\V_C\end{matrix}\right]$

Finally, the required result is:

$\boxed{\frac{\lvert Z'_B \rvert + \lvert Z'_C \rvert} {\lvert Z'_N \rvert} = 2.5293}$

I did not perform any calculations by hand as I used MATLAB. The effective use of linear algebra helps me reach the result quite neatly. Hence my inclination towards it.