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If a+b+c=0,then the family of lines 3ax + by + 2c = 0 pass through the fixed point
Dividing the whole eqn. by 2 and comparing with a+b+c = 03ax + by + 2c = 0 we get,3a/2 x + b/2 y + c = 0 3x/2 = 1x = 2/3andy/2 = 1y = 2.hence (x,y) that is (2/3 , 2) is the fixed point .
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