Apply KCL at node a-

$\dfrac{V_a-240}{3}+$  $\dfrac{V_a-V_b}{6 } +10 = 0$

$3V_a - V_b = 420 \cdots(1)$    

Apply KCL at node b-

$\dfrac{V_b-V_a}{6}$ $+ \dfrac{V_b}{30}$ $+ \dfrac{V_b-V_c}{12} = 0$

$-10V_a + 17V_b - 5V_c = 0 \cdots (2)$

 

• Node c is directly connected to a voltage source of 60 v. …