If a + b + c = 0,

Given:

a + b + c = 0

Substituting c = − a − b in 3ax + by + 2c = 0, we get:

3ax + by – 2a – 2b = 0

⇒ a (3x – 2) + b (y – 2) = 0

This line is of the form L₁ + λL₂ = 0,

which passes through the intersection of the lines L₁ and L_2, i.e. 3x – 2 = 0 and y – 2 = 0 .

Solving 3 x – 2 = 0 and y – 2 = 0, we get:

Hence, the required fixed point is