Answer:  Applying KVL to mesh 1 , 

  $V_1=3I_1+2I_2-I_3$.......................(1)

Applying KVL to mesh 2 ,

$V_2=2I_1+6I_2+4I_3$......................(2)

Applying KVL to mesh 3 , 

$13I_3-I_1+4I_2=0$

$\therefore I_3=\dfrac{1}{13}I_1-\dfrac{4}{13}I_2$......................(3)

Substituting the equation (3) in (1),

$V_1=3I_1+2I_2-\dfrac{1}{3}I_1+\dfrac{4}{13}I_2$

$\therefore V_1=\dfrac{38}{13}I_1+\dfrac{30}{13}I_2$......................(4)

Substituting the equation (3) in (2),

$V_2=2I_1+6I_2+4(\dfrac{1}{13}I_1-\dfrac{4}{13}I_2)$

$\therefore V_2=\dfrac{30}{13}I_1+\dfrac{62}{13}I_2$.......................(5)

 The T network consists of three impedances .

Applying KVL to mesh 1,

$V_1=(Z_ A+Z_C)I_1+Z_CI_2$..............(6)

Applying KVL to mesh 2 ,

$V_2=Z_CI_1+(Z_B+Z_C)I_2$...............(7)

Comparing equations (4) and (5) with (6) and (7),

$Z_A+Z_C=\dfrac{38}{13}$

$Z_C=\dfrac{30}{13}$

$Z_B+Z_C=\dfrac{62}{13}$

Solving the above equations,

$Z_A=\dfrac{8}{13}\Omega$

$Z_B=\dfrac{32}{13}\Omega$

$Z_C=\dfrac{30}{13}\Omega$