# If a + b + c = 0, then the family of lines 3ax + by + 2c = 0 pass through fixed pointA. (2. 2/3)B. (2/3, 2)C. (-2, 2/3)D. none of these

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 L2, 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

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