Max Power

The three parallel AC voltage sources with source impedance is equivalent to a single voltage source $V_S$ with a source impedance $Z_S$ where

$\begin{aligned} V_S & = 10 \angle 0^\circ || 10 \angle 0^\circ || 10 \angle 0^\circ \\ & = 10 \angle 0^\circ \text{ V} \\ Z_S & = (1 +j0)|| (1 +j1)|| (0 +j1)|| \\ & = \dfrac 1{1 + \frac 1{1+j} + \frac 1j} \\ & = \dfrac 1{1+\frac {1-j}2 - j} \\ & = \dfrac 1{\frac 32(1-j)} \\ & = \frac {1+j}3 \ \Omega \end{aligned}$

For maximum active power dissipated by the load the load impedance must be the conjugate of the source impedance. $Z_L = \overline {Z_S} = \dfrac {1-j}3 \ \Omega$ .

Then the power dissipated at the load

$\begin{aligned} P_L & = \Re \left(\left(\frac {Z_L}{Z_S+Z_L}\times V_S\right)^2 \times \frac 1{Z_L}\right) \\ & = \Re \left(\frac {Z_L V_S^2}{(Z_S+Z_L)^2}\right) \\ & = \Re\left(\frac {\frac {1-j}3\times 10^2}{\left(\frac 23\right)^2} \right) \\ & = \Re \left(75(1-j)\right) \\ & = \boxed {75} \text{ W} \end{aligned}$

@Chew-Seong Cheong provided a nice solution which utilizes the conjugate impedance matching principle. For further justification of this approach see the this note