If you plug in point B, you get:
2(a+2b)-1(a+3b)=a+b
2a+4b-a-3b=a+b
a+b=a+b
Therefore, point B will be on any version of this line, regardless of what a&b actually are.
The straight lines x(a+2b) + y(a + 3b) = a + b for different values of a & b pass through a fixed point whose coordinates are :
A)(1,-2)
B)(2, -1)
C)(1, 2)
D)None
Only one option is correct.
It seems that the question is actually an inverted form of finding the locus of the fixed point for some condition. I am unable to make any headway.