Mesh Analysis on AC circuit
You wrote your \$I_3\$ equation for the third mesh incorrectly; it should read \$I_3=\color{red}-10\angle0^\circ\,\text{A}\$. Solving this system of equations yields \$I_1\approx-2.4805-5.4297j\approx5.9695\angle-114.553^\circ\$ and so \$I_0=-I_1\approx5.9695\angle65.447^\circ\$. Note that \$I_0\$ is a phasor; for the time-varying instantaneous current we write \$i_0(t)=\Re\{I_0e^{i\omega t}\}\approx5.9695\cos(\omega t+65.447^\circ)\$.